viết đề cho rõ coi

Có ai giúp tôi với😭😭😭😭😭

Có: \(\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}=\frac{Ax^{2\: }+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(C-B\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+1\right)}\)

Đồng nhất với phân thức \(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)

Ta được: \(\begin{cases}A+B=1\\C-B=2\\A-C=-1\end{cases}\)\(\Leftrightarrow\begin{cases}A=1-B\\C-B=2\\1-B-C=-1\end{cases}\)

\(\Leftrightarrow\begin{cases}A=1-B\\C-B=2\\B+C=2\end{cases}\)\(\Leftrightarrow\begin{cases}A=1-B\\B=0\\C=2\end{cases}\)\(\Leftrightarrow\begin{cases}A=1\\B=0\\C=2\end{cases}\)

\(VP=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A\left(x^2+1\right)}{\left(x-1\right)\left(x^2+1\right)}+\frac{\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\frac{A\left(x^2+1\right)+\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}\)\(=\frac{Ax^2+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\frac{A\left(x^2+1\right)+Bx\left(x-1\right)+C\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)\(=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\frac{Ax^2+A+Bx+C}{x^2+1}\). Lại có: \(VT=\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}=\frac{x-1}{x^2+1}\)

\(\Leftrightarrow\frac{Ax^2+A+Bx+C}{x^2+1}=\frac{x-1}{x^2+1}\Leftrightarrow Ax^2+A+Bx+C=x-1\)

thôi cạn ý tưởng lm tiếp t đi chơi

(ax+b)(x2−x−1)=ax3+cx2+1(ax+b)(x2−x−1)=ax3+cx2+1

⇔ax3+(b−a)x2+(−b−a)x−b=ax3+cx2+0.x+1⇔ax3+(b−a)x2+(−b−a)x−b=ax3+cx2+0.x+1

sử dụng đồng nhất thức ta được: \hept⎧⎨⎩b−a=c−b−a=0−b=1⇔\hept⎧⎨⎩a=1b=−1c=−2