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Ta có: \(VT=(z^2-z+1)(az^2+bz+c)\)
\(=az^4+(b-a)z^3+(c-b+a)z^2+(b-c)z+c\)
Và \(VP=2z^4-z^3+2z^2+1\)
ĐỒng nhất 2 đa thức ta có:
\(\hept{\begin{cases}a=2;b-a=-1\\c-b+a=2\\b-c=z\end{cases}}\) thay lần lượt vào nhé
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)
\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)
c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)
d, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
Mình làm ở đây rồi!
Câu hỏi của Phạm Minh Khôi - Toán lớp 6 - Học toán với OnlineMath
Ta thấy VT = \(az^4+bz^3+cz^2-az^3-bz^2-cz+az^2+bz+c\)
\(=az^4+\left(b-a\right)z^3+\left(c-b+a\right)z^2+\left(b-c\right)z+c\)
Cân bằng hệ số ta có: \(\hept{\begin{cases}a=2\\c=1\end{cases}}\) và \(\hept{\begin{cases}b-a=-1\\c-b+a=2\\b-c=0\end{cases}}\)
Từ đó suy ra \(\hept{\begin{cases}a=2\\b=1\\c=1\end{cases}}\)
Chúc em học tốt :))