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(x-1)(x+1)(x+2)
=(x^2-1)(x+2)
=x^3+2x^2-x-2
[X-1/2] [X+1/2] [4X-1]
=\(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)\)
=\(4x^3-x^2-x+\frac{1}{4}\)
1/2X2Y2 [2X+Y] [2X-Y]
=\(\frac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
=\(2x^2y^2-\frac{1}{2}x^2y^4\)
b) (1 + 2x)(1- 2x) - x(x+2)(x-2)
= (1- 4x2) - x(x2 - 4)
= 1 - 4x2- x3- 4x
= (1 - x3) + (4x - 4x2)
= (1- x) (1 + x + x2) + 4x(1 -x)
= (1-x)(1+5x + x2)
e) Ta có: x4−2x3+2x−1x4−2x3+2x−1
=(x4−1)−2x(x2−1)=(x4−1)−2x(x2−1)
=(x2+1)(x−1)(x+1)−2x(x−1)(x+1)=(x2+1)(x−1)(x+1)−2x(x−1)(x+1)
=(x−1)(x+1)⋅(x2−2x+1)=(x−1)(x+1)⋅(x2−2x+1)
=(x+1)⋅(x−1)3=(x+1)⋅(x−1)3
h) Ta có: 3x2−3y2−2(x−y)23x2−3y2−2(x−y)2
=3(x2−y2)−2(x−y)2=3(x2−y2)−2(x−y)2
=3(x−y)(x+y)−2(x−y)2=3(x−y)(x+y)−2(x−y)2
=(x−y)(3x+3y−2x+2y)=(x−y)(3x+3y−2x+2y)
=(x−y)(x+5y)=(x−y)(x+5y)
Ta có:
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x-y+x\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x\right)-y^2z^2\left(y-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)+z^2\left(z-x\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y^2x+y^2z-z^2y-z^2x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
Ta có: \(\left(x+x^2\right)^{2+1}=0\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]^3=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
x(x-y)+y(x+y)
=x^2-xy+xy+y^2
=x^2+y^2
X[X - Y] + Y[X + Y]
= x2-xy+xy+y2
= x2+y2
X[X2 - Y] - X2 [X + Y] + Y[X2 - Y]
=x3-xy-x3 -x2y+ x2y-y2
= -xy-y2
~ chúc bạn học tốt ~