Phân tích đa thức thành nhân tử:
a) \(x^2-25-4xy+4y^2\)
b) \(x^2-8x+15\)
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\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)
c: \(=\left(5x-y\right)\left(5x+y\right)\)
e: \(=\left(x-2\right)\left(x-3\right)\)
a) x(4y-10x)
b)3(x+2y)+(x+1)
c)(5x-y)(5x+y)
d)5x(y-z)2
e)(x-3)(x-2)
f)(2x+y)3
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
\(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)
`a)x^3-8x^2+16x`
`=x(x^2-8x+16)`
`=x(x-4)^2`
`b)x^2+4y^2+2x-4y-4xy-24`
`=(x-2y)^2+2(x-2y)-24`
`=(x-2y)^2-4(x-2y)+6(x-2y)-24`
`=(x-2y-4)(x-2y+6)`
`c)x^4+x^3-x^2-2x-2`
`=x^4-2x^2+x^3-2x+x^2-2`
`=x^2(x^2-2)+x(x^2-2)+x^2-2`
`=(x^2-2)(x^2+x+1)`
a) \(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\left(x-2y+5\right)\)
b) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
a)\(x^2-25-4xy+4y^2\Leftrightarrow\left(x^2-4xy+4y^2\right)-25\)
\(\Leftrightarrow\left(x-2y\right)^2-5^2\)
\(\Leftrightarrow\left(x-2y-5\right)\left(x-2y+5\right)\)
b)\(x^2-8x+15\Leftrightarrow\left(x-3\right)\left(x-5\right)\)