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

Answer 1

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

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

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

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

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

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

\(\Rightarrow x=\pm2\)

Answer 2

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 \(8x^2-2x=a\Rightarrow64x^2-16x=8a\)

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

\(\Leftrightarrow8a^2+a-9=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{4}\end{matrix}\right.\)