Question

giải pt \(\sqrt{x}+\sqrt{3x-2}=x^2+1\)

Answer 1

Lời giải:

ĐK: $x\geq \frac{2}{3}$

PT \(\sqrt{x}-1+\sqrt{3x-2}=x^2-1\)

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

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

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

\(\Leftrightarrow (x-1)\left[(x-1)+\frac{x-1}{2(\sqrt{x}+1)^2}+\frac{9(x-1)}{2(\sqrt{3x-2}+1)^2}\right]=0\)

\(\Leftrightarrow (x-1)^2\left[1+\frac{1}{2(\sqrt{x}+1)^2}+\frac{9}{2(\sqrt{3x-2}+1)^2}\right]=0\)

Dễ thấy biểu thức trong ngoặc vuông dương, nên $(x-1)^2=0$

$\Rightarrow x=1$ (tm)

Vậy......