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Khách
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MN
1
25 tháng 11 2021
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
KK
24 tháng 1 2017
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
31 tháng 12 2016
\(Q=\frac{2013}{1+x+xy}+\frac{2013}{1+y+yz}+\frac{2013}{1+z+zx}\)
\(=2013\left(\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+xyzx}\right)\)
\(=2013\left(\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+xy+x}\right)=2013\)
XX
0
TD
1
12 tháng 7 2016
Ta có : \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)
\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{1+xz+z}{1+xz+z}=1\)
Bài này phân tích hết ra.
Đặt \(A=\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+xz+1}\)
\(=\frac{\left(x+xy+1\right)\left(y+yz+1\right)+\left(x+xy+1\right)\left(z+xz+1\right)+\left(y+yz+1\right)\left(z+xz+1\right)}{\left(x+xy+1\right)\left(y+yz+1\right)\left(z+xz+1\right)}\)
Đặt \(M=\left(x+xy+1\right)\left(y+yz+1\right)+\left(x+xy+1\right)\left(z+xz+1\right)+\left(y+yz+1\right)\left(z+xz+1\right)\)
\(=\left(x+y+2xy+yz+xy^2+xyz+xy^2z+1\right)+\left(2xz+x+z+xyz+x^2yz+x^2z+xy+1\right)+\left(y+z+2yz+yz^2+xz+xyz+xyz^2+1\right)\)
Thay \(xyz=1;\)có :
\(M=\left(x+y+2xy+yz+xy^2+1+y.1+1\right)+\left(2xz+x+z+1+x.1+x^2z+xy+1\right)+\left(y+z+2yz+yz^2+xz+1+z.1+1\right)\)
\(=3x+3y+3z+3xy+3yz+3xz+xy^2+x^2z+yz^2+6\)
Đặt \(N=\left(x+xy+1\right)\left(y+yz+1\right)\left(z+xz+1\right)\)
\(=\left(x+y+2xy+yz+xy^2+1+y.1+1\right)\left(z+xz+1\right)\)
\(=\left(x+2y+2xy+yz+xy^2+2\right)\left(z+xz+1\right)\)
\(=xz+x^2z+x+3yz+2xy+2y+4xyz+2x^2yz+2xy+yz^2+xyz^2+xy^2z+x^2y^2z+xy^2+2z+2\)
Lần lượt thay \(xyz=1\); cuối cùng có :
\(N=3x+3y+3z+3xy+3yz+3xz+xy^2+x^2z+yz^2+6\)
\(\Rightarrow M=N\)
\(\Rightarrow A=\frac{M}{N}=1\)
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