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Question Number 198814 by hardmath last updated on 24/Oct/23

Simplify: ((((√(x + 4 (√(x - 4)))) + (√(x - 4 (√(x - 4))))) (√(x - 8)))/(4 (√(x^2 - 12x + 32))))

$$\mathrm{Simplify}: \\ $$$$\frac{\left(\sqrt{\mathrm{x}\:+\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\:+\:\sqrt{\mathrm{x}\:-\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\right)\:\sqrt{\mathrm{x}\:-\:\mathrm{8}}}{\mathrm{4}\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{12x}\:+\:\mathrm{32}}} \\ $$

Answered by a.lgnaoui last updated on 24/Oct/23

=(1/4)(√((x+4(√(x−4)))/(x−4))) +(1/4)(√((x−(√(x−4)))/(x−4))) =(((√(x+4(√(x−4)) )) +(√(x−4(√(x−4)))))/( 4(√(x−4)))) =((√((x−4)+4(√(x−4)) +4))/(4(√(x−4))))+((√(x−4)−4(√(x−4)) +4))/(4(√(x−4)))) ((√(((√(x−4)) +2)^2 ))/(4(√(x−4))))+((√(((√(x−4)) )−2)^2 ))/(4(√(x−4)))) =(((√(x−4)) +2)/(4(√(x−4))))+(((√(x−4)) −2)/(4(√(x−4)))) =(1/4)+(1/4)=(1/2)

$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\mathrm{x}+\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}{\boldsymbol{\mathrm{x}}−\mathrm{4}}}\:+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\boldsymbol{\mathrm{x}}−\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}{\boldsymbol{\mathrm{x}}−\mathrm{4}}} \\ $$$$=\frac{\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:}\:+\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}}{\:\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}} \\ $$$$=\frac{\sqrt{\left(\boldsymbol{\mathrm{x}}−\mathrm{4}\right)+\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:+\mathrm{4}}}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}+\frac{\sqrt{\left.\boldsymbol{\mathrm{x}}−\mathrm{4}\right)−\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:+\mathrm{4}}}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}} \\ $$$$\frac{\sqrt{\left(\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:+\mathrm{2}\right)^{\mathrm{2}} }}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}+\frac{\sqrt{\left.\left(\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:\right)−\mathrm{2}\right)^{\mathrm{2}} }}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}} \\ $$$$=\frac{\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:+\mathrm{2}}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}+\frac{\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}\:−\mathrm{2}}{\mathrm{4}\sqrt{\boldsymbol{\mathrm{x}}−\mathrm{4}}}\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by hardmath last updated on 28/Oct/23

thanks professor

$$\mathrm{thanks}\:\mathrm{professor} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Oct/23

((((√(x + 4 (√(x - 4)))) + (√(x - 4 (√(x - 4))))) (√(x - 8)))/(4 (√(x^2 - 12x + 32)))) ((((√(x + 4 (√(x - 4)))) + (√(x - 4 (√(x - 4))))) (√(x - 8)))/(4 (√((x−4)(x−8))))) x>8⇒x−4>4⇒(√(x−4)) >2 (((√(x + 4 (√(x - 4)))) + (√(x - 4 (√(x - 4)))))/(4 (√(x−4)))) Let (√(x - 4)) =y x>8⇒y=(√(x−4)) >2 x=y^2 +4 (((√( y^2 +4+ 4y)) + (√(y^2 +4 - 4y)))/(4y)) ((∣y+2∣+∣y−2∣)/(4y)) ∵ y>2 ∴ ((∣y+2∣−∣y−2∣)/(4y))=(((y+2)+(y−2))/(4y))=((2y)/(4y))=(1/2)

$$\frac{\left(\sqrt{\mathrm{x}\:+\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\:+\:\sqrt{\mathrm{x}\:-\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\right)\:\sqrt{\mathrm{x}\:-\:\mathrm{8}}}{\mathrm{4}\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{12x}\:+\:\mathrm{32}}} \\ $$$$\frac{\left(\sqrt{\mathrm{x}\:+\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\:+\:\sqrt{\mathrm{x}\:-\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\right)\:\sqrt{\mathrm{x}\:-\:\mathrm{8}}}{\mathrm{4}\:\sqrt{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}−\mathrm{8}\right)}} \\ $$$$\:\:\mathrm{x}>\mathrm{8}\Rightarrow\mathrm{x}−\mathrm{4}>\mathrm{4}\Rightarrow\sqrt{\mathrm{x}−\mathrm{4}}\:>\mathrm{2} \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}\:+\:\sqrt{\mathrm{x}\:-\:\mathrm{4}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}}}{\mathrm{4}\:\sqrt{\mathrm{x}−\mathrm{4}}} \\ $$$${Let}\:\sqrt{\mathrm{x}\:-\:\mathrm{4}}\:=\mathrm{y}\:\: \\ $$$$\mathrm{x}>\mathrm{8}\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}−\mathrm{4}}\:>\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{y}^{\mathrm{2}} +\mathrm{4} \\ $$$$\frac{\sqrt{\:\mathrm{y}^{\mathrm{2}} +\mathrm{4}+\:\mathrm{4y}}\:+\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{4}\:-\:\mathrm{4y}}}{\mathrm{4y}} \\ $$$$\frac{\mid\mathrm{y}+\mathrm{2}\mid+\mid\mathrm{y}−\mathrm{2}\mid}{\mathrm{4y}} \\ $$$$\because\:\mathrm{y}>\mathrm{2} \\ $$$$\therefore\:\frac{\mid\mathrm{y}+\mathrm{2}\mid−\mid\mathrm{y}−\mathrm{2}\mid}{\mathrm{4y}}=\frac{\left(\mathrm{y}+\mathrm{2}\right)+\left(\mathrm{y}−\mathrm{2}\right)}{\mathrm{4y}}=\frac{\mathrm{2y}}{\mathrm{4y}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by hardmath last updated on 28/Oct/23

thanks professor

$$\mathrm{thanks}\:\mathrm{professor} \\ $$

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