Question-152993

Question Number 152993 by 7770 last updated on 03/Sep/21

Commented by otchereabdullai@gmail.com last updated on 08/Sep/21

$$\mathrm{nice} \\ $$

Answered by bobhans last updated on 04/Sep/21

(dy/dx)= ((3(√(x^2 +5))−3x((x/( (√(x^2 +5))))))/(x^2 +5)) (dy/dx)=((3x^2 +15−3x^2 )/((x^2 +5)^(3/2) ))=((15)/((x^2 +5)^(3/2) ))

$$\frac{{dy}}{{dx}}=\:\frac{\mathrm{3}\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}−\mathrm{3}{x}\left(\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\right)}{{x}^{\mathrm{2}} +\mathrm{5}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{15}−\mathrm{3}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{15}}{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$ \\ $$

Answered by bobhans last updated on 04/Sep/21

(2)∫_0 ^2 ((3x)/( (√(x^2 +5)))) dx =(3/2)∫_(√5) ^3 (du/( (√u))) =(3/2)[ 2(√u) ]_(√5) ^3 = 3(√3)−3(5)^(1/4)

$$\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\:{dx}\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\sqrt{\mathrm{5}}} ^{\mathrm{3}} \frac{{du}}{\:\sqrt{{u}}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\left[\:\mathrm{2}\sqrt{{u}}\:\right]_{\sqrt{\mathrm{5}}} ^{\mathrm{3}} \:=\:\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{5}}\: \\ $$

Answered by hoochhoch last updated on 04/Sep/21

Commented by benhamimed last updated on 04/Sep/21

x=(√(19−8(√3)))=(√((4−(√3))^2 ))=4−(√3) A=(((19−8(√3))(19−8(√3)−6(4−(√3))−2)+18(4−(√3))+24)/(19−8(√3)−8(4−(√3))+15)) A=(((19−8(√3))(−7−2(√3))+96−18(√3))/2) =((−133+48+56(√3)−38(√3)+96−18(√3))/2) A=((11)/2)

$${x}=\sqrt{\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}}=\sqrt{\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{4}−\sqrt{\mathrm{3}} \\ $$$${A}=\frac{\left(\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}\right)\left(\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{6}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)−\mathrm{2}\right)+\mathrm{18}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)+\mathrm{24}}{\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}−\mathrm{8}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)+\mathrm{15}} \\ $$$${A}=\frac{\left(\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}\right)\left(−\mathrm{7}−\mathrm{2}\sqrt{\mathrm{3}}\right)+\mathrm{96}−\mathrm{18}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{133}+\mathrm{48}+\mathrm{56}\sqrt{\mathrm{3}}−\mathrm{38}\sqrt{\mathrm{3}}+\mathrm{96}−\mathrm{18}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${A}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$

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