Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
làm a thôi nha :D
a) \(C=\left(\frac{x^2+x}{x^2-2x+1}\right):\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{2-x^2}{x\left(x+1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{x+2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{\left(x-1\right)\left(x+1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x+1}{x^2-2x+1}.\frac{x^2-1+x+2-x^2}{x-1}\)
\(C=\frac{x+1}{\left(x^2-2x+1\right)}.\frac{1.x}{x-1}\)
\(C=\frac{\left(x+1\right)^2}{x^3-x^2-2x^2+2x+x-1}\)
\(C=\frac{x^2+2x+1}{x^3-3x^2+3x-1}\)
a)\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1}{x.\left(x-1\right)}+\frac{x}{x.\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1+x-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x.\left(x-1\right)}\right]=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right].\left[\frac{x.\left(x-1\right)}{x+1}\right]=\frac{x.\left(x+1\right).x}{\left(x-1\right).\left(x+1\right)}=\frac{x^2}{x-1}\)
b)\(\text{Để B nguyên }\Rightarrow x^2⋮x-1\)
\(x^2=x^2-1+1=\left(x-1\right).\left(x+1\right)+1\)
\(\Rightarrow\text{Để }x^2⋮x-1\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\Rightarrow x\in\left\{2;0\right\}\)
