Question

\(\sqrt{x+3}+\sqrt{2\cdot x+4}=12-\sqrt{3\cdot x+7}\)

Answer 1

mình bó tay rồi

Answer 2

Đk:\(x\ge-2\)

\(pt\Leftrightarrow\sqrt{x+3}+\sqrt{2x+4}-12+\sqrt{3x+7}=0\)

\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{2x+4}-4+\sqrt{3x+7}-5=0\)

\(\Leftrightarrow\frac{x+3-9}{\sqrt{x+3}+3}+\frac{2x+4-16}{\sqrt{2x+4}+4}+\frac{3x+7-25}{\sqrt{3x+7}+5}=0\)

\(\Leftrightarrow\frac{x-6}{\sqrt{x+3}+3}+\frac{2\left(x-6\right)}{\sqrt{2x+4}+4}+\frac{3\left(x-6\right)}{\sqrt{3x+7}+5}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{1}{\sqrt{x+3}+3}+\frac{2}{\sqrt{2x+4}+4}+\frac{3}{\sqrt{3x+7}+5}\right)=0\)

Dễ thấy:\(\forall x\ge2\) thì \(\frac{1}{\sqrt{x+3}+3}+\frac{2}{\sqrt{2x+4}+4}+\frac{3}{\sqrt{3x+7}+5}>0\) (loại)

Nên \(x-6=0\Rightarrow x=6\) (thỏa)