\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)

\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)

\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)

tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn

cảm ơn cậu giúp mk câu c với ạ

bấm máy tính là ra 1,25 bạn nhé

<=> \(\frac{-2x-1}{12}\)-\(\frac{2x+2}{3}=\frac{1-2x}{4}-\frac{3x-1}{12}\)

<=>\(\frac{-2x-1-8x-8-3+6x+3x-1}{12}=0\)

<=> -x-13=0=> x=-13

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

<=>\(6.\left(\frac{1}{2}-\frac{x+2}{3}\right)-8.\left(x+1\right)=3\left(1-2x\right)-6.\left(\frac{1}{3}-\frac{1-x}{2}\right)\)

<=>3-2.(x+2)-8x-8=3-6x-2+3.(1-x)

<=>3-2x-4-8x-8=3-6x-2+3-3x

<=>-10x-9=-9x+4

<=>x=-13

\(\frac{a}{x}+\frac{b}{x-1}+\frac{c}{x-2}=\frac{9x^2-16x+4}{x^3-3x^2+2x}\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)a}{\left(x-2\right)\left(x-1\right)x}+\frac{\left(x-2\right)xb}{\left(x-2\right)\left(x-1\right)x}+\frac{\left(x-1\right)xc}{\left(x-2\right)\left(x-1\right)x}=\frac{9x^2-16x+4}{x^3-3x^2+2}\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)a+\left(x-2\right)xb+\left(x-1\right)xc}{\left(x-2\right)\left(x-1\right)x}=\frac{9x^2-16x+4}{x^3-3x^2+2}\)

\(\Leftrightarrow\frac{a\left(x^2-3x+2\right)+b\left(x^2-2x\right)+c\left(x^2-x\right)}{x^3-3x^2+2}=\frac{9x^2-16x+4}{x^3-3x^2+2}\)

\(\Leftrightarrow\frac{x^2\left(a+b+c\right)-x\left(3a+2b+c\right)+2a}{x^3-3x^2+2}=\frac{9x^2-16x+4}{x^3-3x^2+2}\)

Sử dụng đồng nhất thức ta được: \(\begin{cases}x^2\left(a+b+c\right)=9\\x\left(3a+2b+c\right)=16\\2a=4\end{cases}\)\(\Leftrightarrow\begin{cases}a=2\\b=3\\c=4\end{cases}\)