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Ta có:\(P=x^3\left(z-y^2\right)+y^3x-y^3z^2+z^3y-z^3x^2+x^2y^2z^2-xyz\)
\(\Rightarrow P=x^3\left(z-y^2\right)+x^2y^2z^2-x^2z^3-\left(y^3z^2-z^3y\right)+y^3x-xyz\)
\(\Rightarrow P=x^3\left(z-y^2\right)+x^2z^2\left(y^2-z\right)-yz^2\left(y^2-z\right)+xy\left(y^2-z\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3-yz^2+xy\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3+xy-yz^2\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2\left(z^2-x\right)+y\left(x-z^2\right)\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2\left(z^2-x\right)-y\left(z^2-x\right)\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(z^2-x\right)\left(x^2-y\right)\)
\(\Rightarrow P=abc\)
Vì a, b, c là hằng số nên P có giá trị không phụ thuộc vào x, y, z
Anh có cách khác nè :
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{yz\left(x-y+z-x\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(yz-xy\right)-\left(z-x\right)\left(zx-yz\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{y\left(x-y\right)\left(z-x\right)-z\left(x-y\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-x\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2z+yz^2-z^2x+zx^2-x^2y+xy^2}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2\left(z-x\right)-zx\left(z-x\right)+y\left(z^2-x^2\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-x\right)\left(yz+xy-y^2-zx\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-y\right)\left[y\left(x-y\right)-z\left(x-y\right)\right]}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?