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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ x3 + x2 z + y2 z - xyz + y3
= (x + y)(x2 - xy + y2) + z(x2 - xy + y2)
= (x2 - xy + y2)(x + y + z)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(a+b\right)+\left(a+b\right)^2\)
\(=\left(a+b\right)\left(1+a+b\right)\)
\(b,4\left(x-y\right)+3\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+3\left(x-y\right)\right)\)
\(=\left(x-y\right)\left(4+3x-3y\right)\)
\(c,\left(a-b\right)+\left(b-a\right)^2\)
\(=\left(a-b\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(1+a-b\right)\)
a) \(\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(1+a+b\right)\)
b) \(4\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left[4+3.\left(x-y\right)\right]\)
c) \(\left(a-b\right)+\left(b-a\right)^2=\left(a-b\right)+\left(b-a\right)\left(b-a\right)\)
\(=\left(a-b\right)-\left(a-b\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(1-b+a\right)\)
d) \(\left(a-b\right)-\left(b-a\right)^2\)
\(=\left(a-b\right)-\left(b-a\right)\left(b-a\right)\)
\(=\left(a-b\right)+\left(a-b\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(1+b-a\right)\)
e) \(a\left(a-b\right)^2-\left(b-a\right)^3\)
\(=a\left(a-b\right)-\left(a-b\right)\left(b-a\right)^2\)
\(=\left(a-b\right)\left[a-\left(b-a\right)^2\right]\)
f) \(\left(y+z\right)\left(12x^2+6x\right)+\left(y-z\right)\left(12x^2+6x\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=\left(12x^2+6x\right)2y\)