Câu hỏi của Phương Thùy Lê

Question

Giải phương trình:

$2x+\left(x+1\right)\sqrt{2x-x^2}=\frac{3}{2}$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $0\le x\le2$

$\Leftrightarrow2x+\left(1-x\right)\left(x+1\right)-\frac{3}{2}+\left(x+1\right)\left(\sqrt{2x-x^2}-\left(1-x\right)\right)=0$

$\Leftrightarrow\frac{-2x^2+4x-1}{2}+\frac{\left(x+1\right)\left(-2x^2+4x-1\right)}{\sqrt{2x-x^2}+1-x}=0$

$\Leftrightarrow\left(-2x^2+4x-1\right)\left(\frac{1}{2}+\frac{x+1}{\sqrt{2x-x^2}+1-x}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}-2x^2+4x-1=0\\frac{x+1}{\sqrt{2x-x^2}+1-x}=-\frac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}-2x^2+4x-1=0\x+3=-\sqrt{2x-x^2}\left(vn\right)\end{matrix}\right.$

$\Leftrightarrow...$Câu trả lời: ĐKXĐ: $0\le x\le2$

$\Leftrightarrow2x+\left(1-x\right)\left(x+1\right)-\frac{3}{2}+\left(x+1\right)\left(\sqrt{2x-x^2}-\left(1-x\right)\right)=0$

$\Leftrightarrow\frac{-2x^2+4x-1}{2}+\frac{\left(x+1\right)\left(-2x^2+4x-1\right)}{\sqrt{2x-x^2}+1-x}=0$

$\Leftrightarrow\left(-2x^2+4x-1\right)\left(\frac{1}{2}+\frac{x+1}{\sqrt{2x-x^2}+1-x}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}-2x^2+4x-1=0\\frac{x+1}{\sqrt{2x-x^2}+1-x}=-\frac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}-2x^2+4x-1=0\x+3=-\sqrt{2x-x^2}\left(vn\right)\end{matrix}\right.$

$\Leftrightarrow...$

§1. Đại cương về phương trình