phân tích đa thức thành nhân tử:
a.\(x^3\left(y-z\right)+y^3\left(z-x\right)+z^3\left(x-y\right)\)
b.\(x^3\left(z-y^2\right)+y^3\left(x-z^2\right)+z^3\left(y-z^2\right)+xyz\left(xyz-1\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]\)
\(=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)\)
\(=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)\)
Cô nghĩ phân tích đa thức này sẽ đẹp hơn:
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)\)
\(=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
b) \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)
Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Hay \(A=3\cdot2x\cdot2y\cdot2z\)
\(A=24xyz\)