Question

mọi người giúp mình phần 1 với ạ!

 

Answer 1

ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(x^2+5x-6+\left(\sqrt{2x+7}-3\right)+\left(\sqrt{3x-2}-1\right)=0\)

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

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)