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a)
\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)
b)
x khác 1
c)
x khác 0; x khác 5
d) x khác 5 ; x khác -5
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
a) ĐKXĐ : \(x\ne\pm2\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{x^2-4}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x-23\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x^2-17x-46\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-\left(3x^2-17x-46\right)-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-3x^2+17x+46-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{2x^2+7x+6}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(2x+3\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+3}{x-2}\)
b) x2 - 1 = 0 <=> x2 = 1 <=> x = ±1
Với x = 1
\(B=\frac{2\cdot1+3}{1-2}=-5\)
Với x = -1
\(B=\frac{2\cdot\left(-1\right)+3}{\left(-1\right)-2}=-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,Đkxđ:x\ne\pm2\)
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+1\right)^2}{x^2-4}\)
b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)
Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)
\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)
Vậy ............