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Question-76122

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Question Number 76122 by benjo last updated on 24/Dec/19
Answered by john santuy last updated on 24/Dec/19
i think this problem not difficult
$${i}\:{think}\:{this}\:{problem}\:{not}\:{difficult} \\ $$
Commented by $@ty@m123 last updated on 24/Dec/19
The term “difficult” is rational. It varies from person to person.
$${The}\:{term}\:“{difficult}''\:\:{is}\:{rational}. \\ $$$${It}\:{varies}\:{from}\:{person}\:{to}\:{person}. \\ $$
Commented by john santuy last updated on 24/Dec/19
yes sir. thanks you
$${yes}\:{sir}.\:{thanks}\:{you} \\ $$
Answered by MJS last updated on 24/Dec/19
x+4−(3/(x−4))=2(√((x+4)/(x−4))) ((x^2 −19)/(x−4))=2(√((x+4)/(x−4))) squaring ⇒ not all solutions will fit the given equation (((x^2 −19)/(x−4)))^2 =4((x+4)/(x−4)) (x^2 −19)^2 =4(x^2 −16) x^4 −42x^2 +425=0 x^2 =25∨x^2 =17 ⇒ x=±5∨x=±(√(17)) testing in given equation ⇒ x=5∨x=−(√(17))
$${x}+\mathrm{4}−\frac{\mathrm{3}}{{x}−\mathrm{4}}=\mathrm{2}\sqrt{\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}}} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{19}}{{x}−\mathrm{4}}=\mathrm{2}\sqrt{\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}}} \\ $$$$\mathrm{squaring}\:\Rightarrow\:\mathrm{not}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{will}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation} \\ $$$$\left(\frac{{x}^{\mathrm{2}} −\mathrm{19}}{{x}−\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{4}\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{19}\right)^{\mathrm{2}} =\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{16}\right) \\ $$$${x}^{\mathrm{4}} −\mathrm{42}{x}^{\mathrm{2}} +\mathrm{425}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mathrm{25}\vee{x}^{\mathrm{2}} =\mathrm{17} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{5}\vee{x}=\pm\sqrt{\mathrm{17}} \\ $$$$\mathrm{testing}\:\mathrm{in}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\Rightarrow\:{x}=\mathrm{5}\vee{x}=−\sqrt{\mathrm{17}} \\ $$
Commented by benjo last updated on 24/Dec/19
thanks you sir
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by behi83417@gmail.com last updated on 24/Dec/19
((x+4)/(x−4))=t^2 ⇒((2x)/8)=((t^2 +1)/(t^2 −1))⇒x=4((t^2 +1)/(t^2 −1)) 4((t^2 +1)/(t^2 −1))+4−2t=(3/(4((t^2 +1)/(t^2 −1))−4))=((3(t^2 −1))/(4(2))) ((8t^2 )/(t^2 −1))−2t=((3(t^2 −1))/8) 64t^2 −16t(t^2 −1)=3(t^2 −1)^2 ⇒3t^4 +16t^3 −70t^2 −16t+3=0 ⇒t=−8.123,−(1/3),+0.123,+3.0 { ((((x+4)/(x−4))=66⇒(x/4)=((67)/(65))⇒x=4.123)),((((x+4)/(x−4))=(1/9)⇒(x/4)=−((10)/8)⇒x=−5.0)),((((x+4)/(x−4))=0.015⇒(x/4)=((1.015)/(0.985))⇒x=−4.123)),((((x+4)/(x−4))=9⇒(x/4)=((10)/8)⇒x=+5.0)) :}
$$\frac{\mathrm{x}+\mathrm{4}}{\mathrm{x}−\mathrm{4}}=\mathrm{t}^{\mathrm{2}} \Rightarrow\frac{\mathrm{2x}}{\mathrm{8}}=\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\Rightarrow\mathrm{x}=\mathrm{4}\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{4}\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}+\mathrm{4}−\mathrm{2t}=\frac{\mathrm{3}}{\mathrm{4}\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}−\mathrm{4}}=\frac{\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}\left(\mathrm{2}\right)} \\ $$$$\frac{\mathrm{8t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}−\mathrm{2t}=\frac{\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{8}} \\ $$$$\mathrm{64t}^{\mathrm{2}} −\mathrm{16t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3t}^{\mathrm{4}} +\mathrm{16t}^{\mathrm{3}} −\mathrm{70t}^{\mathrm{2}} −\mathrm{16t}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}=−\mathrm{8}.\mathrm{123},−\frac{\mathrm{1}}{\mathrm{3}},+\mathrm{0}.\mathrm{123},+\mathrm{3}.\mathrm{0} \\ $$$$\begin{cases}{\frac{\mathrm{x}+\mathrm{4}}{\mathrm{x}−\mathrm{4}}=\mathrm{66}\Rightarrow\frac{\mathrm{x}}{\mathrm{4}}=\frac{\mathrm{67}}{\mathrm{65}}\Rightarrow\mathrm{x}=\mathrm{4}.\mathrm{123}}\\{\frac{\mathrm{x}+\mathrm{4}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{9}}\Rightarrow\frac{\mathrm{x}}{\mathrm{4}}=−\frac{\mathrm{10}}{\mathrm{8}}\Rightarrow\mathrm{x}=−\mathrm{5}.\mathrm{0}}\\{\frac{\mathrm{x}+\mathrm{4}}{\mathrm{x}−\mathrm{4}}=\mathrm{0}.\mathrm{015}\Rightarrow\frac{\mathrm{x}}{\mathrm{4}}=\frac{\mathrm{1}.\mathrm{015}}{\mathrm{0}.\mathrm{985}}\Rightarrow\mathrm{x}=−\mathrm{4}.\mathrm{123}}\\{\frac{\mathrm{x}+\mathrm{4}}{\mathrm{x}−\mathrm{4}}=\mathrm{9}\Rightarrow\frac{\mathrm{x}}{\mathrm{4}}=\frac{\mathrm{10}}{\mathrm{8}}\Rightarrow\mathrm{x}=+\mathrm{5}.\mathrm{0}}\end{cases} \\ $$
Answered by mind is power last updated on 24/Dec/19
⇒1−(2/( (√((x+4)(x−4)))))=(3/((x−4)(x+4))),x>0≠4 x<0,x≠−4 ⇒1+(2/( (√((x−4)(x+4)))))=(3/((4+x)(x−4))), Z=(1/( (√((x−4)(x+4))))) Quadratique
$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}−\mathrm{4}\right)}}=\frac{\mathrm{3}}{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)},\mathrm{x}>\mathrm{0}\neq\mathrm{4} \\ $$$$\mathrm{x}<\mathrm{0},\mathrm{x}\neq−\mathrm{4} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)}}=\frac{\mathrm{3}}{\left(\mathrm{4}+\mathrm{x}\right)\left(\mathrm{x}−\mathrm{4}\right)},\:\:\mathrm{Z}=\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)}} \\ $$$$\mathrm{Quadratique} \\ $$

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