\(\left(x-2\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\1-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)

Vậy phương trình có 2 nghiệm \(x=2;x=\frac{1}{2}\)

\(\left(x-2\right)\left(1-2x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\1-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;\frac{1}{2}\right\}\)

Áp dụng \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\) ta được:

\(a^3-b^3=9\)\(\Leftrightarrow\left(a-b\right)^3+3ab\left(a-b\right)=9\)

Với \(a-b=3\)ta có: \(3^3+3ab.3=9\)

\(\Leftrightarrow27+9ab=9\)\(\Leftrightarrow9ab=18\)\(\Leftrightarrow ab=2\)

Vậy \(ab=2\)

VP của phương trình đâu bạn.

a, \(ĐKXĐ:x\ne2\)

\(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)

\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)

\(\Rightarrow1+3x-6=3-x\)

\(\Leftrightarrow1+3x-6-3+x=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\left(ktm\right)\)

vậy x thuộc tập hợp rỗng

b, \(ĐKXĐ:x\ne\pm1\)

\(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)

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

\(\Rightarrow x^2+x-2x=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x-1=0\Rightarrow x=1\left(ktm\right)\end{cases}}\)

vậy x = 0

c, \(ĐKXĐ:x\ne\pm\frac{1}{2}\)

\(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)

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

\(\Leftrightarrow\frac{32x^2}{12\left(1-2x\right)\left(2x+1\right)}=\frac{-8x\left(2x+1\right)}{12\left(1-2x\right)\left(2x+1\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(2x+1\right)}\)

\(\Rightarrow32x^2=-16x^2-8x-3+6x-24x+48x\)

\(\Leftrightarrow48x^2=22x-3\)

\(\Leftrightarrow48x^2-22x+3=0\)

\(M=\frac{x^4-1-x^4+x^2-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\).\(\frac{x^4.\left(1+x^2\right)}{1+x^2}+\frac{1-x^4}{1+x^2}\)

\(M=\frac{x^2-2}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\).\(\frac{x^4+x^6+1-x^4}{1+x^2}\)

\(M=\frac{x^2-2}{x^6+1}\).\(\frac{x^6+1}{1+x^2}\)

M= \(\frac{x^2-2}{x^2+1}\)

b) M=\(\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)

Tự làm tiếp nhé.