Câu hỏi của Ánh Dương

Question

Giải phương trình sau:

a) $\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3x^2+16}{x^2+10}=1$

b) $2x\left(8x-1\right)^2\left(4x-1\right)=9$

Nguyễn Việt Lâm · Accepted Answer

$\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3\left(x^2+10\right)-14}{x^2+10}-1=0$

$\Leftrightarrow\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}+\frac{14}{x^2+10}-4=0$

$\Leftrightarrow\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1+\frac{14}{x^2+10}-1=0$

$\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0$

$\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0$

$\Leftrightarrow4-x^2=0$ (do $\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0$)

$\Rightarrow x=\pm2$Câu trả lời: $\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3\left(x^2+10\right)-14}{x^2+10}-1=0$

$\Leftrightarrow\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}+\frac{14}{x^2+10}-4=0$

$\Leftrightarrow\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1+\frac{14}{x^2+10}-1=0$

$\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0$

$\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0$

$\Leftrightarrow4-x^2=0$ (do $\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0$)

$\Rightarrow x=\pm2$

Nguyễn Việt Lâm · Answer

b/

$2x\left(4x-1\right)\left(8x-1\right)^2=9$

$\Leftrightarrow\left(8x^2-2x\right)\left(64x^2-16x+1\right)-9=0$

Đặt $8x^2-2x=a\Rightarrow64x^2-16x=8a$

$a\left(8a+1\right)-9=0$

$\Leftrightarrow8a^2+a-9=0$

$\Rightarrow\left[{}\begin{matrix}a=1\a=-\frac{9}{8}\end{matrix}\right.$ $\Rightarrow\left[{}\begin{matrix}8x^2-2x-1=0\8x^2-2x+\frac{9}{8}=0\left(vn\right)\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\x=-\frac{1}{4}\end{matrix}\right.$Câu trả lời: b/