Phân tích đa thức thành nhân tử:
a) \(a+2\sqrt{ab}+b\)
b) \(x^2+2xy+y^2+x^2-y^2\)
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\(a,=x\left(x^2-4x+4-z^2\right)=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-z-2\right)\left(x+z-2\right)\\ b,=\left(x-y\right)^2-\left(z-5\right)^2=\left(x-y-z+5\right)\left(x-y+z-5\right)\)
\(x^3-4x^2+4x-xz^2=x\left(x^2-4x+4-z^2\right)\)
\(=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-2-z\right)\left(x-2+z\right)\)
\(x^2-2xy+y^2-z^2+10z-25\)
\(=\left(x-y\right)^2-\left(z-5\right)^2\)
\(=\left(x-y+z-5\right)\left(x-y-z+5\right)\)
a) \(x^2-xy+x-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-y\right)\)
b) \(x^2+2xy-4x-8y\)
\(=x\left(x+2y\right)-4\left(x+2y\right)\)
\(\left(x-4\right)\left(x+2y\right)\)
c) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a:=2(x-2)+y(x-2)
=(x-2)(y+2)
b: \(=\left(x+y\right)^2-4\)
=(x+y+2)(x+y-2)
\(a,=5x\left(x-1\right)+y\left(x-1\right)=\left(5x+y\right)\left(x-1\right)\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
a)5x2-5x+xy-y=(5x2-5x)+(xy-y)=5x(x-1)+y(x-1)=(x-1)(5x+y)
b)x2-2xy+y2-9=(x2-2xy+y2)-9=(x-y)2-32=(x-y-3)(x-y+3)
a. 16xy2 - 12xy + 24x2y
= 4xy(4y - 3 + 6x)
c. 16 - x2 + 2xy - y2
= 42 - (x2 - 2xy + y2)
= 42 - (x - y)2
= (4 - x + y)(4 + x - y)
b: \(x^3-x^2-x+1=\left(x-1\right)^2\left(x+1\right)\)
d: \(x^2-x-20=\left(x-5\right)\left(x+4\right)\)
`a)(x+2)^2+2(x^2-4)+(x-2)^2`
`=(x+2)^2+2(x-2)(x+2)+(x-2)^2`
`=(x+2+x-2)^2=(2x)^2=4x^2`
`b)x^2-x+1/4`
`=x^2-2.x .1/2+1/4=(x-1/2)^2`
`c)(x+y)^3-(x-y)^3`
`=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]`
`=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)`
`=2y(3x^2+y^2)`
a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)
\(=\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+2+x-2\right)^2=\left(2x\right)^2=4x^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
c) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)
a, \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)
b,\(x^2+2xy+y^2+x^2-y^2=\left(x+y\right)^2+\left(x-y\right)\left(x+y\right)\)\(=\left(x+y\right)\left(x+y+x-y\right)=2x\left(x+y\right)\)