Question

a) \(\sqrt{x+10}-\sqrt{3+x}=\sqrt{4x-23}\)

b) \(\sqrt{3x+4}-\sqrt{1+2x}=\sqrt{x+3}\)

Answer 1

a.

ĐKXĐ: \(x\ge\dfrac{23}{4}\)

\(\sqrt{x+10}=\sqrt{x+3}+\sqrt{4x-23}\)

\(\Leftrightarrow x+10=5x-20+2\sqrt{\left(x+3\right)\left(4x-23\right)}\)

\(\Leftrightarrow\sqrt{4x^2-11x-69}=15-2x\) \(\left(x\le\dfrac{15}{2}\right)\)

\(\Leftrightarrow4x^2-11x-69=\left(15-2x\right)^2\)

\(\Leftrightarrow49x-294=0\)

\(\Leftrightarrow x=6\) (thỏa mãn)

Answer 2

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

\(\sqrt{3x+4}=\sqrt{2x+1}+\sqrt{x+3}\)

\(\Leftrightarrow3x+4=3x+4+2\sqrt{\left(2x+1\right)\left(x+3\right)}\)

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

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\left(loại\right)\end{matrix}\right.\)