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Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)
\(E=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{xyz.z}{zx+xyz.z+xyz}=\frac{1}{yz+y+1}+\frac{y}{yz+y+1}+\frac{yz}{1+yz+y}=\frac{1+y+yz}{1+y+yz}=1\)
Xin lỗi mk chưa học tới bài này.Bạn vào câu hỏi tương tự thử có k.
\(M=\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-zx}\)
Đặt \(N=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Vậy \(M=\frac{N}{x^2+y^2+z^2-xy-yz-zx}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2+y^2+z^2-xy-yz-zx}=x+y+z=2016\)
(*) bn ghi sai đề 1 chỗ nhé:ở mẫu thức của M phải là \(x^2+y^2+z^2-xy-yz-zx\) nhé!
x^2+y^2+z^2=xy+yz+xz
=>2x^2+2y^2+2z^2-2xy-2yz-2xz=0
=>(x-y)^2+(y-z)^2+(x-z)^2=0
=>x=y=z
=>A=(x-x)^2022+(y-y)^2023=0
\(A=\frac{\left(xy+2016z\right)\left(yz+2016x\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)
Thay \(x+y+z=2016\)
\(A=\frac{\left[xy+\left(x+y+z\right)z\right]\left[yz+\left(x+y+z\right)x\right]\left[zx+\left(x+y+z\right)y\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)
\(A=\frac{\left[xy+xz+yz+z^2\right]\left[yz+xy+xz+x^2\right]\left[zx+xy+yz+y^2\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left[x\left(y+z\right)+z\left(y+z\right)\right]\left[y\left(z+x\right)+x\left(z+x\right)\right]\left[x\left(z+y\right)+y\left(z+y\right)\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left[\left(y+z\right)\left(x+z\right)\right]\left[\left(x+z\right)\left(x+y\right)\right]\left[\left(z+y\right)\left(x+y\right)\right]}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left(x+z\right)\left(x+z\right)\left(y+z\right)\left(y+z\right)\left(x+y\right)\left(x+y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=\frac{\left(x+z\right)^2\left(y+z\right)^2\left(x+y\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(x+z\right)^2}\)
\(A=1\)