phân tích đa thức sau thành nhân tử:
\(A=10x^4-27x^3y-110x^2y^2-27x^3y+10y^4\)
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a: Ta có: \(10x^4-27x^3y-110x^2y^2-27xy^3+10y^4\)
\(=10x^4+20x^2y^2+10y^4-27xy\left(x^2+y^2\right)-130x^2y^2\)
\(=10\left(x^2+y^2\right)^2-27xy\left(x^2+y^2\right)-130x^2y^2\)
\(=10\left(x^2+y^2\right)^2-52xy\left(x^2+y^2\right)+25xy\left(x^2+y^2\right)-130x^2y^2\)
\(=2\left(x^2+y^2\right)\left(5x^2+5y^2-26xy\right)+5xy\left(5x^2+5y^2-26xy\right)\)
\(=\left(5x^2-26xy+5y^2\right)\left(2x^2+5xy+2y^2\right)\)
\(=\left(5x^2-25xy-xy+5y^2\right)\left(2x^2+4xy+xy+2y^2\right)\)
\(=\left\lbrack5x\left(x-5y\right)-y\left(x-5y\right)\right\rbrack\left\lbrack2x\left(x+2y\right)+y\left(x+2y\right)\right\rbrack\)
=(5x-y)(x-5y)(2x+y)(x+2y)
b: \(x^5-4x^4+3x^3+3x^2-4x+1\)
\(=x^5+x^4-5x^4-5x^3+8x^3+8x^2-5x^2-5x+x+1\)
\(=\left(x+1\right)\left(x^4-5x^3+8x^2-5x+1\right)\)
\(=\left(x+1\right)\left(x^4-x^3-4x^3+4x^2+4x^2-4x-x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x^3-4x^2+4x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left\lbrack\left(x^3-x^2\right)-3x^2+3x+x-1\right\rbrack\)
\(=\left(x+1\right)\left(x-1\right)\cdot\left(x-1\right)\left(x^2-3x+1\right)=\left(x+1\right)\left(x-1\right)^2\cdot\left(x^2-3x+1\right)\)
1. = (3x)3 - (ab)3
= (3x - ab)[(3x)2 + 3x . ab + ab2)
= (3x - ab)(9x2 + 3xab + ab2)
\(3z^2+6zy+3y^2-27x^2\)
\(=3\left(z^2+2zy+y^2-9x^2\right)\)
\(=3\left(\left(z+y\right)^2-\left(3x\right)^2\right)\)
\(=3\left(z+y-3x\right)\left(z+y+3x\right)\)
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
d)5.(x-y)-y(x-y)
=(x-y)(5-y)
e) y.(x-z)+7(z-x)
=y.(x-z)-7(x-z)
=(x-z)(y-7)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)