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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)\left(x-3\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x-2\right)\left(x^2-9\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3-2x^2-9x+18-x^3-3x^2-3x-1\)
\(=-5x^2-12x+17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)^3-\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)\\ =\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x^3-1\right)\\ =2.\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-\left(x^3-1\right)\\ =2.\left(3x^2+1\right)-\left(x^3-1\right)\\ =6x^2+2-x^3+1=-x^3+6x^2+3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x +1)3(x-1)+x3-3x(x+1)(x-1)
=(x3+3x2+3x+1)(x-1)+x3-3x(x2-1)
=x4-x3+3x3-3x2+3x2-3x+x-1+x3-3x3+3x
=x4+x-1
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Lời giải:
ĐKXĐ: $x\neq 2; x\neq -3$
\(\frac{x+2}{x+3}-\frac{5}{(x+3)(x-2)}-\frac{1}{x-2}=\frac{(x+2)(x-2)-5-(x+3)}{(x+3)(x-2)}\\ =\frac{x^2-4-5-x-3}{(x+3)(x-2)}=\frac{x^2-x-12}{(x+3)(x-2)}\\ =\frac{(x+3)(x-4)}{(x+3)(x-2)}=\frac{x-4}{x-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 3)(x + 3) - x(x - 1)
= x² - 9 - x² + x
= (x² - x²) + x - 9
= x - 9