Câu hỏi của Phan uyển nhi

Question

Giải phương trình sau :

$\sqrt{x}+\sqrt{2x-1}+x^2+x-4=\text{0}$

Hồng Phúc · Accepted Answer

ĐK: $x\ge\dfrac{1}{2}$

$pt\Leftrightarrow\sqrt{x}-1+\sqrt{2x-1}-1+x^2+x-2=0$

$\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2x-2}{\sqrt{2x-1}+1}+\left(x-1\right)\left(x+2\right)=0$

$\Leftrightarrow\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2\right)\left(x-1\right)=0$

Vì $\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2>0$ nên $x-1=0\Leftrightarrow x=1\left(tm\right)$

Câu trả lời:

ĐK: $x\ge\dfrac{1}{2}$

$pt\Leftrightarrow\sqrt{x}-1+\sqrt{2x-1}-1+x^2+x-2=0$

$\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2x-2}{\sqrt{2x-1}+1}+\left(x-1\right)\left(x+2\right)=0$

$\Leftrightarrow\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2\right)\left(x-1\right)=0$

Vì $\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2>0$ nên $x-1=0\Leftrightarrow x=1\left(tm\right)$