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\(-3x^2y:xy=-3\left(x^2y:xy\right)=-3x\)
\(2a^3b^3:\left(-2ab^2\right)=\left(2:-2\right)\left(a^3:a\right)\left(b^3:b^2\right)=-a^2b\)
\(\dfrac{1}{5}m^2n^3:\left(-5m^2n^2\right)=\left(\dfrac{1}{5}:-5\right)\left(m^2:m^2\right)\left(n^3:n^2\right)=-\dfrac{1}{25}n\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(2a^2-3b-6a+ab=a\left(2a+b\right)-3\left(2a+b\right)=\left(a-3\right)\left(2a+b\right)\)
\(\left(a-b\right)^2-4=\left(a-b\right)^2-2^2=\left(a-b-2\right)\left(a-b+2\right)\)
\(4x^2-12xy+3x-9y=4x\left(x-3y\right)+3\left(x-3y\right)=\left(x-3y\right)\left(4x+3\right)\)
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a) \(3a-3b+a^2-2ab+b^2\)
\(=3\left(a-b\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a-b+3\right)\)
a)
3.(a-b) +2.(a-b ) =5 .(a-b )
câu b làm tương tự nha nhóm a^2 -2ab +b^2 vào 1nhoms và làm như câu a
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)
\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)
\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)
\(=\dfrac{3a-1}{2}\)
\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)
b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)
\(=\dfrac{1}{b-3}\)
\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)
\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)
{b} ^ {2} + 2ab + \ frac {1} {3} {b} ^ {3} +3 {B} ^ {2}NS2+2MộtNS+3NS3+3NS2
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1
Đơn giản hóa \ frac {1} {3} {b} ^ {3}31NS3 đến \ frac {{b} ^ {3}} {3}3NS3.
{b} ^ {2} + 2ab + \ frac {{b} ^ {3}} {3} +3 {B} ^ {2}NS2+2MộtNS+3+3NS2