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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{-3+6}{9x^2-4}=\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2_{ }-3x+2-3}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{\left(3x-2\right)\left(3x+2\right)}\)
ý b bạn chỉ cần phân tích mẩu ra nó là 1 hằng đẳng thức rồi lấy mẫu chung là hằng đẩng thưcs đó rồi làm như thường
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x}{x+3}+\dfrac{2-x}{x+3}\\ =\dfrac{x+2-x}{x+3}\\ =\dfrac{2}{x+3}\\b,\dfrac{x^2y}{x-y}-\dfrac{xy^2}{x-y}\\ =\dfrac{x^2y-xy^2}{x-y}\\ =\dfrac{xy\left(x-y\right)}{x-y}\\ =xy\\ c,\dfrac{2x}{2x-y}+\dfrac{y}{y-2x}\\=\dfrac{2x}{2x-y}-\dfrac{y}{2x-y}\\ =\dfrac{2x-y}{2x-y}\\ =1 \)
`a, x/(x+3) + (2-x)/(x+3) = (x+2-x)/(x+3) = 2/(x+3)`
`b, (x^2y)/(x-y) - (xy^2)/(x-y) = (x^2y-xy^2)/(x-y) = (xy(x-y))/(x-y)= xy`
`c, (2x)/(2x-y) - (y)/(2x-y)`
`= (2x-y)/(2x-y) = 1`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x-2\right)\left(15x-7y\right)\\ b,=x\left(x-11\right)\left(2x-1\right)\\ c,=2x\left(x-3\right)\left(2+3y\right)\\ d,=\left(x-y\right)\left(x-7y\right)\\ e,=\left(x-3\right)\left(4x-12-2x\right)\\ =\left(x-3\right)\left(2x-12\right)=2\left(x-6\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x-3}-\frac{3}{2x+6}-\frac{x}{2x^2-12x+18}\)
\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x^2-6x+9\right)}\)
\(=\frac{1}{x-3}-\frac{3}{2\left(x+3\right)}-\frac{x}{2\left(x-3\right)^2}\)
\(=\frac{2\left(x-3\right)\left(x+3\right)-3\left(x-3\right)^2-x\left(x+3\right)}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{2\left(x^2-9\right)-3\left(x^2-6x+9\right)-x\left(x+3\right)}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{2x^2-18-3x^2+18x-27-x^2-3x}{2\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{-2x^2+15x-45}{2\left(x-3\right)^2\left(x+3\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2y\left(x+2\right)-3x-6\)
\(=2y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(2y-3\right)\)
b: \(3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(x+4\right)\left(3-x\right)\)
c: \(2\left(x+5\right)-x^2-4x\)
\(=2x+10-x^2-4x\)
\(=-x^2-2x+10\)
\(=-x^2-2x-1+11\)
\(=11-\left(x^2+2x+1\right)\)
\(=11-\left(x+1\right)^2\)
\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)
d: \(x^2+6x-3x-18\)
\(=\left(x^2+6x\right)-\left(3x+18\right)\)
\(=x\left(x+6\right)-3\left(x+6\right)\)
\(=\left(x+6\right)\left(x-3\right)\)
Ta có: