Cho \(xyz=1\) . Giá trị của biểu thức \(Q=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\) bằng ?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2015x}{xy+2015x+2015}+\frac{y}{yz+y+2015}+\frac{z}{xz+z+1}\)
Thay 2015=xyz vào A, ta được
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz+xy+xyz}{xy\left(xz+z+1\right)}=\frac{xy\left(xz+1+z\right)}{xy\left(xz+z+1\right)}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nhầm xíu nhá mk lm lại :
\(A=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)\(=\frac{xz}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)
\(=\frac{xz+z+1}{xz+z+1}=1\)
\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xy}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}=\frac{xy}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xy+1+z}{xz+z+1}=1\)
vậy A=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(xy+yz+xz=\frac{2^2-18}{2}=-7\)
\(x+y+z=2\)=> \(z-1=-x-y+1\)
=> \(\frac{1}{xy+z-1}=\frac{1}{xy-x-y+1}=\frac{1}{\left(x-1\right)\left(y-1\right)}\)
Tương tự \(\frac{1}{yz+x-1}=\frac{1}{\left(y-1\right)\left(z-1\right)};\frac{1}{xz+y-1}=\frac{1}{\left(z-1\right)\left(x-1\right)}\)
=> \(S=\frac{x+y+z-3}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=-\frac{1}{xyz-\left(yz+xy+xz\right)+\left(x+y+z\right)-1}\)
\(=\frac{-1}{-1+7+2-1}=-\frac{1}{7}\)
Vậy \(S=-\frac{1}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Từ x+y=1
=>x=1-y
Ta có: \(x^3+y^3=\left(1-y\right)^3+y^3=1-3y+3y^2-y^3+y^3\)
\(=3y^2-3y+1=3\left(y^2-y+\frac{1}{3}\right)=3\left(y^2-2.y.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)\)
\(=3\left[\left(y-\frac{1}{2}\right)^2+\frac{1}{12}\right]=3\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\) với mọi y
=>GTNN của x3+y3 là 1/4
Dấu "=" xảy ra \(< =>\left(y-\frac{1}{2}\right)^2=0< =>y=\frac{1}{2}< =>x=y=\frac{1}{2}\) (vì x=1-y)
Vậy .......................................
b) Ta có: \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{y+x}\)
\(=\left(\frac{x^2}{y+z}+x\right)+\left(\frac{y^2}{z+x}+y\right)+\left(\frac{z^2}{y+z}+z\right)-\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{y+z}-\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}-1\right)\)
Đặt \(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}\)
\(A=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{y+x}+1\right)-3\)
\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{y+x}-3\)
\(=\left(x+y+z\right)\left(\frac{1}{y+x}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\)
\(=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\ge\frac{9}{2}-3=\frac{3}{2}\)
(phần này nhân phá ngoặc rồi dùng biến đổi tương đương)
\(=>P=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}-1\right)\ge2\left(\frac{3}{2}-1\right)=1\)
=>minP=1
Dấu "=" xảy ra <=>x=y=z
Vậy.....................
tich đã