Câu hỏi của super potato

Question

giải pt : $x^2-4x+12=4\sqrt{4-x}+\sqrt{3x+16}$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $-\frac{16}{3}\le x\le4$

$\Leftrightarrow3x^2-12x+36=12\sqrt{4-x}+3\sqrt{3x+16}$

$\Leftrightarrow3x^2-9x+4\left(6-x-3\sqrt{4-x}\right)+\left(x+12-3\sqrt{3x+16}\right)=0$

$\Leftrightarrow3\left(x^2-3x\right)+\frac{4\left(x^2-3x\right)}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+12+3\sqrt{3x+16}}=0$

$\Leftrightarrow\left(x^2-3x\right)\left(3+\frac{4}{6-x+3\sqrt{4-x}}+\frac{1}{x+12+3\sqrt{3x+16}}\right)=0$

$\Leftrightarrow x^2-3x=0$Câu trả lời: ĐKXĐ: $-\frac{16}{3}\le x\le4$

$\Leftrightarrow3x^2-12x+36=12\sqrt{4-x}+3\sqrt{3x+16}$

$\Leftrightarrow3x^2-9x+4\left(6-x-3\sqrt{4-x}\right)+\left(x+12-3\sqrt{3x+16}\right)=0$

$\Leftrightarrow3\left(x^2-3x\right)+\frac{4\left(x^2-3x\right)}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+12+3\sqrt{3x+16}}=0$

$\Leftrightarrow\left(x^2-3x\right)\left(3+\frac{4}{6-x+3\sqrt{4-x}}+\frac{1}{x+12+3\sqrt{3x+16}}\right)=0$

$\Leftrightarrow x^2-3x=0$