cm P=1/x+y+z * 1/xy+yz+zx * (1/y+1/x+1/z)*(1/xy+1/yz+1/zx) luôn dương với mọi giá trị của x,y,z
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25 tháng 6 2016
Sửa lại đề là x;y;z khác -1.
\(A=\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+z+x+1}=\)
\(A=\frac{x\left(y+1\right)+x+1}{x\left(y+1\right)+y+1}+\frac{y\left(z+1\right)+y+1}{y\left(z+1\right)+z+1}+\frac{z\left(x+1\right)+z+1}{z\left(x+1\right)+x+1}=\)
\(A=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\frac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\frac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}=\)vì x;y;z khác -1 nên:
\(A=\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}=\)
\(A=\frac{x}{x+1}+\frac{1}{x+1}+\frac{y}{y+1}+\frac{1}{y+1}+\frac{z}{z+1}+\frac{1}{z+1}=\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}=1+1+1=3\)
A = 3 với mọi x;y;z khác -1 nên A không phụ thuộc vào x;y;z. đpcm
28 tháng 8 2018
\(A=\frac{xy+2y+1}{xy+x+y+1}+\frac{yz+2z+1}{yz+y+z+1}+\frac{zx+2x+1}{zx+z+x+1}\)
\(=\frac{y\left(x+1\right)+y+1}{\left(x+1\right)\left(y+1\right)}+\frac{z\left(y+1\right)+z+1}{\left(y+1\right)\left(z+1\right)}+\frac{x\left(z+1\right)+x+1}{\left(z+1\right)\left(x+1\right)}\)
\(=\frac{y}{y+1}+\frac{1}{x+1}+\frac{z}{z+1}+\frac{1}{y+1}+\frac{x}{x+1}+\frac{1}{z+1}\)
\(=\frac{y+1}{y+1}+\frac{z+1}{z+1}+\frac{x+1}{x+1}=3\)
MN
1
25 tháng 11 2021
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)