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![](https://rs.olm.vn/images/avt/0.png?1311)
a
Ta có
\(2x^2+2x=2x\left(x+1\right)\)
b
\(\left(1+xy\right)^2-\left(x+y\right)^2=\left(1+xy-x-y\right)\left(1+xy+x+y\right)\)
\(\left[\left(1-x\right)-y\left(1-x\right)\right]\left[\left(1+x\right)+y\left(1+x\right)\right]=\left(1-x\right)\left(1-y\right)\left(1+x\right)\left(1+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+8x^2+17x+10\)
\(=x^3+2x^2+x^2+5x^2+10x+5x+2x+10\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(5x^2+5x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+5x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: M = xy(x+y) + yz(y+z) + xz (x+z) + 2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(x + y)
= (x + y)(xy + zx + zy + z2)
= (x + y)[x(y + z) + z(y + z)]
M = (x + y)(y + z)(z + x) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
B = (x + 3)(x - 1)(x - 5)(x + 15) + 64x2
B = x4 + 12x3 - 58x2 - 180x + 225 + 64x2
B = x4 + 12x3 + 6x2 - 180x + 225
![](https://rs.olm.vn/images/avt/0.png?1311)
x4 - x2 + 4x - 1
= x4 - ( x2 - 4x + 1 )
= (x2)2 - ( x - 1 )2
= ( x2 - x +1 ).( x2 +x -1 )
Chúc bạn học tốt nha !
![](https://rs.olm.vn/images/avt/0.png?1311)
(x + y)3 - 1 - 3xy(x + y - 1)
= x3 + 3x2y + 3xy2 + y3 - 1 - 3x2y - 3xy2 + 3xy
= x3 - 1 + 3xy
= x(x2 + 3y) - 1
k bt lm nx r :v
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right) \)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)
= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)
= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)
= \(\left(x^2+x+1\right)\left(x^2+x-2\right)+4\left(x^2+x-2\right)\)
= \(\left(x^2+x+5\right)\left(x^2+x-2\right)\)