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Câu 1: A
Câu 2: B
Câu 3: D
Câu 4: A
Câu 5: C
Câu 6: B
Câu 7: A
Câu 9: B
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\(1,\\ a,\dfrac{8x}{2xy}=\dfrac{4x}{y}\\ b,\dfrac{2xy}{6y}=\dfrac{x}{3}\\ c,\dfrac{3\left(x+2\right)}{2x}=\dfrac{6\left(x+2\right)}{4x}\\ d,\dfrac{4\left(x-2\right)}{3\left(x+1\right)}=\dfrac{8\left(x-2\right)x}{6\left(x+1\right)x}\\ 2,\\ \dfrac{x^2+3x+2}{x^2+x}=\dfrac{x^2+x+2x+2}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)}=\dfrac{x+2}{x}\\ 3,\\ \dfrac{x^2-3x}{x^2-9}=\dfrac{x}{x+3}\)
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Bài 3:
Ta có: \(x^2-2x+4=\left(x-1\right)^2+3\ge3\forall x\)
\(\Leftrightarrow P=\dfrac{15}{x^2-2x+4}=\dfrac{15}{\left(x-1\right)^2+3}\le5\forall x\)
Dấu '=' xảy ra khi x=1
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a) \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{x+1}{x-2}\Leftrightarrow A=x+1\)
b) \(\dfrac{M}{x-1}=\dfrac{x^2+3x+2}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=\dfrac{\left(x+1\right)\left(x+2\right)}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=x+2\Leftrightarrow M=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6: \(=x^3\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x^2+x+1\right)\)
7: =(x-4)(x+2)
2/
\(2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\)
3/
\(9x^2-\left(x-1\right)^2=\left(3x\right)^2-\left(x-1\right)^2=\left(3x-x+1\right)\left(3x+x-1\right)\)
4/
\(x^2-3x+6y-4y^2=x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7: =(x-4)(x+2)
4: \(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-3\right)\)