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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
e: Ta có: \(x\left(y-x\right)^2-x^2+2xy-y^2\)
\(=x\left(x-y\right)^2-\left(x-y\right)^2\)
\(=\left(x-y\right)^2\cdot\left(x-1\right)\)
Bài 2:
a: Ta có: \(M=m^2\left(m+n\right)-n^2m-n^3\)
\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)
\(=\left(m+n\right)^2\cdot\left(m-n\right)\)
\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)
=0
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Ta có \(x^2-\left(m+n\right)x+m.n=\left(x^2-mx\right)-\left(nx-m.n\right)\)
\(=x\left(x-m\right)-n\left(x-m\right)=\left(x-m\right)\left(x-n\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^m+2)+(x^m) = 2xm+2 = 2(xm+1)
(x^x+1)-(x^x)-1 = xx+1-xx-1 = 0
(m^4)-(n^4) = (m2)2-(n2)2 = (m2-n2)(m2+n2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)
\(=xz\left(x+y\right)\left(x-z\right)\)
tìm số nguyên m,n,k nguyên sao cho khi phân tích đa thức (x-10)(x-k)+1 thành nhân tử được (x+m)(x+n)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2\left(m+n\right)-3y^2\left(m+n\right)\)
\(=\left(m+n\right)\left(x^2-3y^2\right)\)
\(=\left(m+n\right)\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
b) \(2x\left(x-4\right)-5\left(4-x\right)-\left(2x+5\right)\left(7-5x\right)\)
\(=2x\left(x-4\right)+5\left(x-4\right)-\left(2x+5\right)\left(7-5x\right)\)
\(=\left(x-4\right)\left(2x+5\right)-\left(2x+5\right)\left(7-5x\right)\)
\(=\left(2x+5\right)\left(x-4-7+5x\right)\)
\(=\left(2x+5\right)\left(6x-11\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(x^3z+x^2yz-x^2z^2-xyz^2=\left(x^3z-x^2z^2\right)+\left(x^2yz-xyz^2\right)\)
\(=\left(x-z\right)\left(x^2z+xyz\right)\)
\(=xz\left(x-z\right)\left(x+y\right)\)
b ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)