b) Ta có: \(x^3-x^2y-xy^2+y^3\)

\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)^2\)

a: \(2x^2+3xy-14y^2\)

\(=2x^2+7xy-4xy-14y^2\)

\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)

\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)

\(=\left(2x+7y\right)\left(x-2y\right)\)

b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)

\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)

\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)

\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)

\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)

c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)

\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)

\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)

\(=\left(7x-5\right)\left(-2x-2\right)\)

\(=-2\left(x+1\right)\left(7x-5\right)\)

d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)

\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)

\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)

\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)

\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)

\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)

\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)

\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)

a) 12x²-3xy+8xz-2yz=3x(4x-y)+2z(4x-y)=(3x+2z)(4x-y)

b) x³+x²y-x²z-xyz=x(x²+xy-xz-yz)=x²(x+y-z-yz)

Phan h cac đa thuc thành nhân tử

Phan h cac đa thuc thanh nhan tử

`@` `\text {Ans}`

`\downarrow`

`a,`

`3x + 6xy + 3y - 3z`

`= 3(x + 2y + y - z)`

`b,`

`x+ xy - xz - xyz`

`= x(1 + y)*(1-z)`

a: 3x^2+6xy+3y^2-3z^2

=3(x^2+2xy+y^2-z^2)

=3[(x+y)^2-z^2]

=3(x+y+z)(x+y-z)

b: x+xy-xz-xyz

=x(y+1)-xz(y+1)

=(y+1)*x*(1-z)

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

=3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

\((x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3\)

\(=-3[x^4y^2-x^4z^2-x^2y^2z^2+x^2z^4-x^2y^4+x^2y^2z^2+y^4z^2-y^2z^4\)

\(=-3[x^2(x^2y^2-x^2z^2-z^2y^2+z^4)-y^2(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2)(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2[x^2(y^2-z^2)-z^2(y^2-z^2)]\)

\(=-3(x^2-y^2)(x^2-z^2)(y^2-z^2)\)

\(=-3(x-y)(x+y)(x-z)(x+z)(y+z)(y-z)\)

a)\(3xy-6y=3y\left(x-2\right)\)

b)\(x\left(x+y\right)+2x+2y=x\left(x+y\right)+\left(2x+2y\right)=x\left(x+y\right)+2\left(x+y\right)=\left(x+y\right)\left(x+2\right)\)

c)\(y^2-81=y^2-9^2=\left(y-9\right)\left(y+9\right)\)

a)`3xy-6y`

`=3y(x-2)`

b)`x(x+y)+2x+2y`

`=x(x+y)+2(x+y)`

`=(x+y)(x+2)`

c)`y^2 -81`

`=y^2-9^2`

`=(y-9)(y+9)`

a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)

\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)

\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

Đặt \(a=x^2-5x+5\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)

\(\Leftrightarrow A=a^2-1^2+1\)

\(\Leftrightarrow A=a^2\)

Thay \(a=x^2-5x+5\)vào A ta có :

\(A=\left(x^2-5x+5\right)^2\)

b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)

\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)

\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)

\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

Làm tương tự câu a)

c) \(12x^2-3xy-8xz+2yz\)

\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)

\(=\left(4x-y\right)\left(3x-2z\right)\)

\(z^3\left(x+y^2\right)+y^3\left(z-x^2\right)-x^3\left(y+z^2\right)-xyz\left(xyz-1\right)\)

\(=xz^3+y^2z^3+y^3z-x^2y^3-x^3-x^3z^2-x^2y^2z^2+xyz\)

\(=\left(y^2z^3+y^3z\right)+\left(xz^3+xyz\right)-\left(x^2y^3+x^2y^2z^2\right)-x^3\left(y+z^2\right)\)

\(=y^2z\left(y+z^2\right)+xz\left(y+z^2\right)-x^2y^2\left(y+z^2\right)-x^3\left(y+z^2\right)\)

\(=\left(y+z^2\right)\left(y^2z+xz-x^2y^2-x^3\right)\)

\(=\left(y+z^2\right)\left[z\left(y^2+x\right)-x^2\left(y^2+x\right)\right]\)

\(=\left(y+z^2\right)\left(z-x^2\right)\left(y^2+x\right)\)

Tick hộ nha bạn 😘

z^3(x+y^2)+y^3(z-x^2)-x^3(y+z^2)-xyz(xyz-1)

Đây nè bạn.

a, \(x^2-6=x^2-\sqrt{6^2}=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

b, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\sqrt{3}=\left(x+\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

c, \(x^2-2\sqrt{5}x+5=x^2-2\sqrt{5}x+\sqrt{5}=\left(x-\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x-\sqrt{5}\right)\)

a: \(x^2-6=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

b: \(x^2+2\sqrt{3}x+3=\left(x+\sqrt{3}\right)^2\)

c: \(x^2-2\sqrt{5}x+5=\left(x-\sqrt{5}\right)^2\)