Question

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\(\left(x+1\right)\sqrt{x+2}+\left(x+6\right)\sqrt{x+7}=x^2+7x+12\)

Answer 1

\(\left(x+1\right)\sqrt{x+2}+\left(x+6\right)\sqrt{x+7}=x^2+7x+12\)

ĐK: \(x\ge -2\)

\(\Leftrightarrow\left(x+1\right)\left(\sqrt{x+2}-2\right)+\left(x+6\right)\left(\sqrt{x+7}-3\right)=x^2+2x-8\)

\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{x+2-4}{\sqrt{x+2}+2}+\left(x+6\right)\cdot\dfrac{x+7-9}{\sqrt{x+7}+3}=\left(x-2\right)\left(x+4\right)\)

\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{x-2}{\sqrt{x+2}+2}+\left(x+6\right)\cdot\dfrac{x-2}{\sqrt{x+7}+3}-\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{x+1}{\sqrt{x+2}+2}+\dfrac{x+6}{\sqrt{x+7}+3}-x-4\right)=0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

Answer 2

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Answer 3