Question

a, \(\dfrac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)

b,\(\left(\sqrt{1+x}-1\right)\left(\sqrt{1-x}+1\right)=2x\)

Answer 1

a. 

ĐKXĐL \(x\ge-\dfrac{1}{3}\)

\(\dfrac{3x}{\sqrt{3x+10}}=\dfrac{3x}{\sqrt{3x+1}+1}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{3x+10}=\sqrt{3x+1}+1\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow3x+10=3x+2+2\sqrt{3x+1}\)

\(\Leftrightarrow\sqrt{3x+1}=4\)

\(\Leftrightarrow x=5\)

Answer 2

b.

ĐKXĐ: \(-1\le x\le1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{\sqrt{1-x}+1}{\sqrt{1+x}+1}=2\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow\sqrt{1-x}+1=2\sqrt{1+x}+2\)

\(\Leftrightarrow\sqrt{1-x}=2\sqrt{1+x}+1\)

\(\Leftrightarrow1-x=4\left(x+1\right)+1+4\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=-5x-4\) (\(x\le-\dfrac{4}{5}\))

\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)