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Hồi lớp 8 mk làm bài này hoài:
Ta có: \(\dfrac{x}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{z}{zx+z+1}\)
\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{xyz+xy+x}+\dfrac{xyz}{x^2yz+xyz+xy}\)
\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{xy+x+1}+\dfrac{1}{xy+x+1}\) ( vì \(xyz=1\) )
\(=\dfrac{x+xy+1}{xy+x+1}\)
\(=1\)
Hok tốt!
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
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 \(yz+xy+xz=xyz\left(x+y+z\right)\)
cũng dễ thôi
M=\(\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)
\(M=\dfrac{z}{z\left(1+x+xy\right)}+\dfrac{xz}{xz\left(1+y+yz\right)}+\dfrac{xyz}{xyz\left(1+z+zx\right)}\\ =\dfrac{z}{z+xz+xyz}+\dfrac{xz}{xz+xyz+xyz\left(z\right)}+\dfrac{xyz}{xyz+xyz\left(z\right)+xyz\left(xz\right)}\\ màxyz=1\\ nênM=\dfrac{z}{z+xz+1}+\dfrac{xz}{z+xz+1}+\dfrac{1}{z+xz+1}\\ vậyM=\dfrac{z+xz+1}{z+xz+1}=1\)
xét tử số:
x+y+z-(xy+yz+zx+1)+xyz
=(y-1)+z-(xy-x)-yz-zx+xyz
=(y-1)-x(y-1)-(yz-z)+(xyz-xz)
=(y-1)-x(y-1)-z(y-1)+xz(y-1)
=(y-1)(1+xz-x-z)
=(y-1)[(xz-z)-(x-1)]
=(y-1)[z(x-1)-(x-1)]
=(y-1)(z-1)(x-1)
Xét mẫu sô:
x^2.y+1-x^2-y
=(x^2.y-y)-(x-1)
=(x^2-1)y-(x-1)
=(x-1)(x+1)(y-1)
Thay tử và mẫu số vào phân thức đại số trên ta được:
\(\dfrac{\left(x-1\right)\left(y-1\right)\left(z-1\right)}{\left(x-1\right)\left(y-1\right)\left(x+1\right)}=\dfrac{z-1}{x+1}\)
Vậy.....
nhó tick cho mình để ủng hộ mình nhé
xin chân thành cảm ơn
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)