Câu hỏi của Thanh Tân

Question

$x^3+10-2\sqrt{2x+1}=2\left(8x+\sqrt{16-3x}\right)$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $-\frac{1}{2}\le x\le\frac{16}{3}$

$\Leftrightarrow x^3-16x+\left(x+2-2\sqrt{2x+1}\right)+\left(-x+8-2\sqrt{16-3x}\right)=0$

$\Leftrightarrow\left(x^2-4x\right)\left(x+4\right)+\frac{x^2-4x}{x+2+2\sqrt{2x+1}}+\frac{x^2-4x}{-x+8+2\sqrt{16-3x}}=0$

$\Leftrightarrow\left(x^2-4x\right)\left(x+4+\frac{1}{x+2+2\sqrt{2x+1}}+\frac{1}{-x+8+2\sqrt{16-3x}}\right)=0$

$\Leftrightarrow x^2-4x=0\Rightarrow\left[{}\begin{matrix}x=0\x=4\end{matrix}\right.$Câu trả lời: ĐKXĐ: $-\frac{1}{2}\le x\le\frac{16}{3}$

$\Leftrightarrow x^3-16x+\left(x+2-2\sqrt{2x+1}\right)+\left(-x+8-2\sqrt{16-3x}\right)=0$

$\Leftrightarrow\left(x^2-4x\right)\left(x+4\right)+\frac{x^2-4x}{x+2+2\sqrt{2x+1}}+\frac{x^2-4x}{-x+8+2\sqrt{16-3x}}=0$

$\Leftrightarrow\left(x^2-4x\right)\left(x+4+\frac{1}{x+2+2\sqrt{2x+1}}+\frac{1}{-x+8+2\sqrt{16-3x}}\right)=0$

$\Leftrightarrow x^2-4x=0\Rightarrow\left[{}\begin{matrix}x=0\x=4\end{matrix}\right.$

Violympic toán 9