Phân tích các đa thức sau thành nhân tử tổng hợp :
a, 5y^3 - 10xy^2 + 5yx^2 - 20y
b, x^2 + 2xy + y^2 - xz - yz
c, 9x^2 + y^2 + 6xy
d, 8 -12x + 6x^2 - x^3
e, 125x^3 - 75x^2 + 15x - 1
Giúp mình với ạ mình đang cần gấp
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do hơi bận nên mk ghi đáp án nha, ko hiểu đâu ib mk
a) \(3xy^2-2xy+12x=x\left(3y^2-2y+12\right)\)
b) \(x^3-10x^2+25x-16xy^2=x\left(x-4y-5\right)\left(x+4y-5\right)\)
c) \(5y^3-10xy^2+5x^2y-20y=5y\left(y-x-2\right)\left(y-x+2\right)\)
d) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)\left(x+y-z\right)\)
e) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
f) \(8-12x+6x^2-x^3=\left(2-x\right)^3\)
g) \(125x^3-75x^2+15x-1=\left(5x-1\right)^3\)
h) \(x^2-xz-9y^2+3yz=\left(x-3y\right)\left(x+3y-z\right)\)
dài quá mình làm 3 câu đầu thôi nhé!
a)7x^2-14xy
=7x(x-2y)
b) 3x^2-6xy+3y^2
=3(x^2-2xy+y^2)
c) x^2-4z^2-2xy+y^2
=(x^2-2xy+y^2)-4z^2
=(x-y-2z)(x-y+2z)
=3(x-y)^2
c)
\(10x\left(x-y\right)-6y\left(y-x\right)\)
\(=10x\left(x-y\right)+6x\left(x-y\right)\)
\(=\left(10x+6x\right)\left(x-y\right)\)
\(c,3x^2+5y-3xy-5x\)
\(=\left(3x^2-3xy\right)+\left(5y-5x\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)
\(5y^3-10xy^2+5yx^2-20y=5y\left(y^2-2xy+x^2-4\right)\)
\(=5y\left[\left(x+y\right)^2-4\right]\)
\(=5y\left(x+y-2\right)\left(x+y+2\right)\)
Chúc bạn học tốt.
x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
***
a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
***
2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
\(\left(x-1\right)^2-25\)
\(=x^2-2x+1-25\)
\(=x^2-2x-24\)
\(=x^2-6x+4x-24\)
\(=x.\left(x-6\right)+4.\left(x-6\right)\)
\(=\left(x+4\right).\left(x-6\right)\)
a, \(1-2y+y^2=\left(y+1\right)^2=\left(y+1\right)\left(y+1\right)\)
b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
c, \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
d, \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
a) Ta có: \(5y^3-10xy^2+5yx^2-20y\)
\(=5y\left(y^2-2xy+x^2-4y\right)\)
b) Ta có: \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\cdot\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
c) Ta có: \(9x^2+y^2+6xy\)
\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)
\(=\left(3x+y\right)^2\)
d) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
e) Ta có: \(125x^3-75x^2+15x-1\)
\(=\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot1+3\cdot5x\cdot1^2-1^3\)
\(=\left(5x-1\right)^3\)