Question

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

b)\(\sqrt{3x+1}+\sqrt{5x+4}=x^2+x+3\)

Answer 1

a.

ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x-2}-1)=0$

$\Leftrightarrow \frac{x+1-4}{\sqrt{x+1}+2}+\frac{x-2-1}{\sqrt{x-2}+1}=0$

$\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow (x-3)\left[\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x-2}+1}\right]=0$

Dễ thấy biểu thức trong ngoặc vuông luôn dương, do đó $x-3=0$

$\Leftrightarrow x=3$ (tm)

Answer 2

b. ĐKXĐ: $x\geq \frac{-1}{3}$

PT \(\Leftrightarrow x^2+x+3-\sqrt{3x+1}-\sqrt{5x+4}=0\)

\(\Leftrightarrow (x^2-x)+[(x+1)-\sqrt{3x+1}]+[(x+2)-\sqrt{5x+4}]=0\)

\(\Leftrightarrow (x^2-x)+\frac{(x+1)^2-(3x+1)}{x+1+\sqrt{3x+1}}+\frac{(x+2)^2-(5x+4)}{x+2+\sqrt{5x+4}}=0\)

\(\Leftrightarrow (x^2-x)+\frac{x^2-x}{x+1+\sqrt{3x+1}}+\frac{x^2-x}{x+2+\sqrt{5x+4}}=0\)

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

Hiển nhiên biểu thức trong ngoặc vuông $>0$ với mọi $x\geq \frac{-1}{3}$

Do đó: $x^2-x=0$
$\Leftrightarrow x(x-1)=0$

$\Leftrightarrow x=0$ hoặc $x=1$ (tm)