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\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=x²(x-3)-4x+3.4
=x²(x-3)-4(x+3)
=x²(x-3)+4(x-3)
=(x-3)(x²+4)
=(x-3)(x²+2²)
=(x-3)(x-2)(x+2)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^2\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-4\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
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Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-4x^2-8x+8=\left(x^2-6x+4\right)\left(x+2\right)\)
Hc tốt
Trả lời:
\(x^3-4x^2-8x+8\)
\(=x^3+2x^2-6x^2-12x+4x+8\)
\(=x^2.\left(x+2\right)-6x.\left(x+2\right)+4.\left(x+2\right)\)
\(=\left(x+2\right).\left(x^2-6x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
bạn làm theo cách này nhen
\(\left(4x+1\right)\left(4x-3\right)\left(2x-3\right)\left(8x+8\right)-130=\left(16x^2-8x-3\right)\left(16x-8x-24\right)-130\)
đặt \(16x^2-8x-3=y< =>y\left(y-21\right)-130=y^2-21y-130=y^2-2.y.\frac{21}{2}+\left(\frac{21}{2}\right)^2-\left(\frac{21}{2}\right)^2-130=\left(y-\frac{21}{2}\right)^2-240,25=\)
\(=\left(y-\frac{21}{2}\right)^2-15,5^2=\left(y-26\right)\left(y+12\right)\)
có chõ nào sai hay bạn không hiểu thì cũng đừng trách nhen
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
c) \(\left(2y-1\right)^1-4x^2+4x-1=\left(2y-1\right)^2-\left(2x-1\right)^2=\left(2y-1-2x+1\right)\left(2y-1+2x-1\right)\)
\(=\left(2y-2x\right)\left(2y+2x-2\right)=4\left(y-x\right)\left(y+x-1\right)\)
\(x^3-4x^2+8x-8=x^2\left(x-2\right)-2x\left(x-2\right)+4\left(x-2\right)=\left(x-2\right)\left(x^2-2x+4\right)\)
\(x^3-4x^2+8x-8\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+4\right)\)