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Question Number 203066 by mathlove last updated on 09/Jan/24
if p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) and x−(4/x)=2 then show that (((32)/p))^2 =80
$${if}\:{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}\:\:\:{and}\:{x}−\frac{\mathrm{4}}{{x}}=\mathrm{2} \\ $$$${then}\:{show}\:{that}\:\:\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\mathrm{80} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jan/24
if p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) and x−(4/x)=2 then show that (((32)/p))^2 =80_ _ •p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) =(x/(1−(2/x)))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) =((x(1+(2/x))(1+(4/x^2 ))−x(1−(2/x))(1+(4/x^2 ))−4(1−(2/x))(1+(2/x)))/((1−(2/x))(1+(2/x))(1+(4/x^2 )))) =(((x+2)(1+(4/x^2 ))+(2−x)(1+(4/x^2 ))−4(1−(2/x))(1+(2/x)))/((1−(4/x^2 ))(1+(4/x^2 )))) =(((1+(4/x^2 ))(x+2+2−x)−4(1−(2/x))(1+(2/x)))/((1−((16)/x^4 )))) =((4(1+(4/x^2 ))−4(1−(4/x^2 )))/((x^4 −16)/x^4 )) =((4(1+(4/x^2 )−1+(4/x^2 )))/((x^4 −16)/x^4 )) =((4((8/x^2 )))/((x^4 −16)/x^4 )) =((32)/x^2 )×(x^4 /(x^4 −16)) =((32x^2 )/(x^4 −16)) =((32)/(x^2 −((16)/x^2 ))) =((32)/((x−(4/x))(x+(4/x)))) =((32)/((2)(x+(4/x)))) p=((16)/(x+(4/x))) • p^2 =((16^2 )/((x+(4/x))^2 ))=((16^2 )/((x−(4/x))^2 +16))=((16^2 )/((2)^2 +16))=((16^2 )/(20)) ▶ (((32)/p))^2 =((32^2 )/p^2 )=((32^2 )/((16^2 )/(20)))=32^2 ×((20)/(16^2 ))=80
$${if}\:{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}\:\:\:{and}\:{x}−\frac{\mathrm{4}}{{x}}=\mathrm{2} \\ $$$$\underline{{then}\:{show}\:{that}\:\:\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\mathrm{80}_{\:_{\:} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:}\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$$$\bullet{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:=\frac{{x}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)−{x}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}{\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)} \\ $$$$\:\:=\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)+\left(\mathrm{2}−{x}\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}{\left(\mathrm{1}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)} \\ $$$$\:\:=\frac{\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)\left({x}+\mathrm{2}+\mathrm{2}−{x}\right)−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)}{\left(\mathrm{1}−\frac{\mathrm{16}}{{x}^{\mathrm{4}} }\right)} \\ $$$$\:\:=\frac{\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)}{\frac{{x}^{\mathrm{4}} −\mathrm{16}}{{x}^{\mathrm{4}} }} \\ $$$$\:\:=\frac{\mathrm{4}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)}{\frac{{x}^{\mathrm{4}} −\mathrm{16}}{{x}^{\mathrm{4}} }} \\ $$$$\:\:=\frac{\mathrm{4}\left(\frac{\mathrm{8}}{{x}^{\mathrm{2}} }\right)}{\frac{{x}^{\mathrm{4}} −\mathrm{16}}{{x}^{\mathrm{4}} }} \\ $$$$\:\:\:=\frac{\mathrm{32}}{{x}^{\mathrm{2}} }×\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{4}} −\mathrm{16}} \\ $$$$\:\:=\frac{\mathrm{32}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} −\mathrm{16}} \\ $$$$\:\:=\frac{\mathrm{32}}{{x}^{\mathrm{2}} −\frac{\mathrm{16}}{{x}^{\mathrm{2}} }} \\ $$$$\:=\frac{\mathrm{32}}{\left({x}−\frac{\mathrm{4}}{{x}}\right)\left({x}+\frac{\mathrm{4}}{{x}}\right)} \\ $$$$\:=\frac{\mathrm{32}}{\left(\mathrm{2}\right)\left({x}+\frac{\mathrm{4}}{{x}}\right)} \\ $$$$\:\:{p}=\frac{\mathrm{16}}{{x}+\frac{\mathrm{4}}{{x}}} \\ $$$$\bullet\:{p}^{\mathrm{2}} =\frac{\mathrm{16}^{\mathrm{2}} }{\left({x}+\frac{\mathrm{4}}{{x}}\right)^{\mathrm{2}} }=\frac{\mathrm{16}^{\mathrm{2}} }{\left({x}−\frac{\mathrm{4}}{{x}}\right)^{\mathrm{2}} +\mathrm{16}}=\frac{\mathrm{16}^{\mathrm{2}} }{\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{16}}=\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{20}} \\ $$$$\blacktriangleright\:\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\frac{\mathrm{32}^{\mathrm{2}} }{{p}^{\mathrm{2}} }=\frac{\mathrm{32}^{\mathrm{2}} }{\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{20}}}=\mathrm{32}^{\mathrm{2}} ×\frac{\mathrm{20}}{\mathrm{16}^{\mathrm{2}} }=\mathrm{80} \\ $$
Answered by Rasheed.Sindhi last updated on 10/Jan/24
x−(4/x)=2⇒ determinant (((x^2 =2x+4))) p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) =((2x+4)/(x−2))−(x/(1+(2/x)))−(4/(1+(4/(2x+4)))) =((2(x+2))/(x−2))−(x^2 /(x+2))−(4/((2x+8)/(2x+4))) =((2(x+2))/(x−2))−((2x+4)/(x+2))−((4(x+2))/(x+4)) =((2(x+2))/(x−2))−((4(x+2))/(x+4))−2 =((2(x+2)(x+4)−4(x+2)(x−2))/((x−2)(x+4)))−2 =((2x^2 +12x+16−4x^2 +16)/(x^2 +2x−8))−2 =((−2x^2 +12x+32)/(x^2 +2x−8))−2 =((−2x^2 +12x+32−2x^2 −4x+16)/(x^2 +2x−8)) =((−4x^2 +8x+48)/(x^2 +2x−8)) =((−4(2x+4)+8x+48)/(2x+4+2x−8)) =((−8x−16+8x+48)/(4x−4))=((32)/(4(x−1)))=(8/(x−1)) ▶(((32)/p))^2 =(((32)/(8/(x−1))))^2 =(((32(x−1))/8))^2 =16(x^2 −2x+1) =16(2x+4−2x+1)=16(5)=80
$${x}−\frac{\mathrm{4}}{{x}}=\mathrm{2}\Rightarrow\begin{array}{|c|}{{x}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{4}}\\\hline\end{array} \\ $$$$\:{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{\mathrm{2}{x}+\mathrm{4}}{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{2}{x}+\mathrm{4}}} \\ $$$$\:\:=\frac{\mathrm{2}\left({x}+\mathrm{2}\right)}{{x}−\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}}−\frac{\mathrm{4}}{\frac{\mathrm{2}{x}+\mathrm{8}}{\mathrm{2}{x}+\mathrm{4}}} \\ $$$$\:\:=\frac{\mathrm{2}\left({x}+\mathrm{2}\right)}{{x}−\mathrm{2}}−\frac{\mathrm{2}{x}+\mathrm{4}}{{x}+\mathrm{2}}−\frac{\mathrm{4}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{2}\left({x}+\mathrm{2}\right)}{{x}−\mathrm{2}}−\frac{\mathrm{4}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{4}}−\mathrm{2} \\ $$$$\:\:=\frac{\mathrm{2}\left({x}+\mathrm{2}\right)\left({x}+\mathrm{4}\right)−\mathrm{4}\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}−\mathrm{2} \\ $$$$\:\:=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{16}−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}−\mathrm{2} \\ $$$$\:\:=\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{32}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}−\mathrm{2} \\ $$$$\:\:=\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{32}−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{16}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}} \\ $$$$\:\:=\frac{−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{48}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}} \\ $$$$\:\:=\frac{−\mathrm{4}\left(\mathrm{2}{x}+\mathrm{4}\right)+\mathrm{8}{x}+\mathrm{48}}{\mathrm{2}{x}+\mathrm{4}+\mathrm{2}{x}−\mathrm{8}} \\ $$$$=\frac{−\mathrm{8}{x}−\mathrm{16}+\mathrm{8}{x}+\mathrm{48}}{\mathrm{4}{x}−\mathrm{4}}=\frac{\mathrm{32}}{\mathrm{4}\left({x}−\mathrm{1}\right)}=\frac{\mathrm{8}}{{x}−\mathrm{1}} \\ $$$$\blacktriangleright\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{32}}{\frac{\mathrm{8}}{{x}−\mathrm{1}}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{32}\left({x}−\mathrm{1}\right)}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{16}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$=\mathrm{16}\left(\mathrm{2}{x}+\mathrm{4}−\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{16}\left(\mathrm{5}\right)=\mathrm{80} \\ $$
Commented by mathlove last updated on 09/Jan/24
ok thanks
$${ok}\:{thanks} \\ $$
Answered by AST last updated on 09/Jan/24
x^2 −4=2x⇒(x−2)(x+2)=2x (x−1)^2 −5=0⇒x=1+_− (√5) ⇒p=(x/(2/(x+2)))−(x/(2/(x−2)))−(4/(((x+4))/((x+2))))=((x(x+2)−x(x−2))/2)−((4(x+2))/(x+4)) =2x−((4(x+2))/(x+4))=((2(x^2 +2x−4))/(x+4))=((8x)/(x+4)) (((32)/p))^2 =[((32(x+4))/(8x))]^2 =[((4(x+4))/x)]^2 =(4+((16)/x))^2 =(+_− 4(√5))^2 =80
$${x}^{\mathrm{2}} −\mathrm{4}=\mathrm{2}{x}\Rightarrow\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)=\mathrm{2}{x} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}=\mathrm{0}\Rightarrow{x}=\mathrm{1}\underset{−} {+}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{p}=\frac{{x}}{\frac{\mathrm{2}}{{x}+\mathrm{2}}}−\frac{{x}}{\frac{\mathrm{2}}{{x}−\mathrm{2}}}−\frac{\mathrm{4}}{\frac{\left({x}+\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)}}=\frac{{x}\left({x}+\mathrm{2}\right)−{x}\left({x}−\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{4}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{4}} \\ $$$$=\mathrm{2}{x}−\frac{\mathrm{4}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{4}}=\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}\right)}{{x}+\mathrm{4}}=\frac{\mathrm{8}{x}}{{x}+\mathrm{4}} \\ $$$$\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\left[\frac{\mathrm{32}\left({x}+\mathrm{4}\right)}{\mathrm{8}{x}}\right]^{\mathrm{2}} =\left[\frac{\mathrm{4}\left({x}+\mathrm{4}\right)}{{x}}\right]^{\mathrm{2}} =\left(\mathrm{4}+\frac{\mathrm{16}}{{x}}\right)^{\mathrm{2}} =\left(\underset{−} {+}\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{80} \\ $$
Answered by AST last updated on 09/Jan/24
p=(x^2 /(x−2))−(x^2 /(x+2))−((4x^2 )/(x^2 +4))=4x^2 ((1/(x^2 −4))−(1/(x^2 +4))) =((32x^2 )/((x^2 −4)(x^2 +4)))=((32x^2 )/(2x(2x+8)))=((8x)/(x+4)) ⇒(((32)/p))^2 =[((4(x+4))/x)]^2 =(+_− 4(√5))^2 =80
$${p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}}=\mathrm{4}{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{32}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{\mathrm{32}{x}^{\mathrm{2}} }{\mathrm{2}{x}\left(\mathrm{2}{x}+\mathrm{8}\right)}=\frac{\mathrm{8}{x}}{{x}+\mathrm{4}} \\ $$$$\Rightarrow\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\left[\frac{\mathrm{4}\left({x}+\mathrm{4}\right)}{{x}}\right]^{\mathrm{2}} =\left(\underset{−} {+}\mathrm{4}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{80} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jan/24
x−(4/x)=2⇒ determinant ((((4/x)=x−2)))⇒ determinant ((((2/x)=((x−2)/2)))) ⇒ determinant (((x^2 =2x+4))) p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) =(x/(1−(2/x)))−(x/(1+(2/x)))−(4/(1+(4/x)∙(1/x))) =(x/(1−((x−2)/2)))−(x/(1+((x−2)/2)))−(4/(1+(x−2)((1/x)))) =(x/((2−x+2)/2))−(x/((2+x−2)/2))−(4/(1+((x−2)/x))) =(x/((4−x)/2))−(x/(x/2))−(4/(1+1−(2/x))) =((2x)/(4−x))−2−(4/(2−((x−2)/2))) =((2x)/(4−x))−2−(4/((6−x)/2)) =((2x)/(4−x))−(8/(6−x))−2 =((12x−2x^2 −32+8x)/(x^2 −10x+24))−2 =((−2x^2 +20x−32−2x^2 +20x−48)/(x^2 −10x+24)) =((−4x^2 +40x−96−80+96)/(x^2 −10x+24))=((−4(2x+4)+40x−80)/(2x+4−10x+24)) =((−8x−16+40x−80)/(−8x+28)) =((32x−96)/(−8x+28)) =((8x−24)/(−2x+7)) ▶(((32)/p))^2 =((32^2 )/p^2 )=((32^2 )/((64(x−3)^2 )/((−2x+7)^2 ))) =((32×32(−2x+7)^2 )/(64(x−3)^2 ))=((16(4x^2 −28x+49))/(x^2 −6x+9)) =((16{4(2x+4)−28x+49})/(2x+4−6x+9))=((16(−20x+65))/(−4x+13))=((16×5(−4x+13))/(−4x+13))=80
$${x}−\frac{\mathrm{4}}{{x}}=\mathrm{2}\Rightarrow\begin{array}{|c|}{\frac{\mathrm{4}}{{x}}={x}−\mathrm{2}}\\\hline\end{array}\Rightarrow\begin{array}{|c|}{\frac{\mathrm{2}}{{x}}=\frac{{x}−\mathrm{2}}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\begin{array}{|c|}{{x}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{4}}\\\hline\end{array} \\ $$$$\:{p}=\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{2}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}−\frac{{x}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{\mathrm{4}}{{x}}\centerdot\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:=\frac{{x}}{\mathrm{1}−\frac{{x}−\mathrm{2}}{\mathrm{2}}}−\frac{{x}}{\mathrm{1}+\frac{{x}−\mathrm{2}}{\mathrm{2}}}−\frac{\mathrm{4}}{\mathrm{1}+\left({x}−\mathrm{2}\right)\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\:\:=\frac{{x}}{\frac{\mathrm{2}−{x}+\mathrm{2}}{\mathrm{2}}}−\frac{{x}}{\frac{\mathrm{2}+{x}−\mathrm{2}}{\mathrm{2}}}−\frac{\mathrm{4}}{\mathrm{1}+\frac{{x}−\mathrm{2}}{{x}}} \\ $$$$\:\:=\frac{{x}}{\frac{\mathrm{4}−{x}}{\mathrm{2}}}−\frac{{x}}{\frac{{x}}{\mathrm{2}}}−\frac{\mathrm{4}}{\mathrm{1}+\mathrm{1}−\frac{\mathrm{2}}{{x}}} \\ $$$$\:\:=\frac{\mathrm{2}{x}}{\mathrm{4}−{x}}−\mathrm{2}−\frac{\mathrm{4}}{\mathrm{2}−\frac{{x}−\mathrm{2}}{\mathrm{2}}} \\ $$$$\:\:=\frac{\mathrm{2}{x}}{\mathrm{4}−{x}}−\mathrm{2}−\frac{\mathrm{4}}{\frac{\mathrm{6}−{x}}{\mathrm{2}}} \\ $$$$\:\:=\frac{\mathrm{2}{x}}{\mathrm{4}−{x}}−\frac{\mathrm{8}}{\mathrm{6}−{x}}−\mathrm{2} \\ $$$$\:\:\:=\frac{\mathrm{12}{x}−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{32}+\mathrm{8}{x}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{24}}−\mathrm{2} \\ $$$$\:\:=\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{32}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{48}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{24}} \\ $$$$\:\:=\frac{−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{40}{x}−\mathrm{96}−\mathrm{80}+\mathrm{96}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{24}}=\frac{−\mathrm{4}\left(\mathrm{2}{x}+\mathrm{4}\right)+\mathrm{40}{x}−\mathrm{80}}{\mathrm{2}{x}+\mathrm{4}−\mathrm{10}{x}+\mathrm{24}} \\ $$$$=\frac{−\mathrm{8}{x}−\mathrm{16}+\mathrm{40}{x}−\mathrm{80}}{−\mathrm{8}{x}+\mathrm{28}} \\ $$$$=\frac{\mathrm{32}{x}−\mathrm{96}}{−\mathrm{8}{x}+\mathrm{28}} \\ $$$$=\frac{\mathrm{8}{x}−\mathrm{24}}{−\mathrm{2}{x}+\mathrm{7}} \\ $$$$\blacktriangleright\left(\frac{\mathrm{32}}{{p}}\right)^{\mathrm{2}} =\frac{\mathrm{32}^{\mathrm{2}} }{{p}^{\mathrm{2}} }=\frac{\mathrm{32}^{\mathrm{2}} }{\frac{\mathrm{64}\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{\left(−\mathrm{2}{x}+\mathrm{7}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{32}×\mathrm{32}\left(−\mathrm{2}{x}+\mathrm{7}\right)^{\mathrm{2}} }{\mathrm{64}\left({x}−\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{16}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{49}\right)}{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$$$=\frac{\mathrm{16}\left\{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{4}\right)−\mathrm{28}{x}+\mathrm{49}\right\}}{\mathrm{2}{x}+\mathrm{4}−\mathrm{6}{x}+\mathrm{9}}=\frac{\mathrm{16}\left(−\mathrm{20}{x}+\mathrm{65}\right)}{−\mathrm{4}{x}+\mathrm{13}}=\frac{\mathrm{16}×\mathrm{5}\cancel{\left(−\mathrm{4}{x}+\mathrm{13}\right)}}{\cancel{−\mathrm{4}{x}+\mathrm{13}}}=\mathrm{80} \\ $$
Commented by mathlove last updated on 10/Jan/24
so good
$${so}\:{good} \\ $$

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