Cho các số thực x,y,z thỏa mãn đồng thời các điều kiện \(x+y+z=2\) , \(x^2+y^2+z^2=18\)và \(xyz=-1\)
Tính giá trị của \(S=\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1}\)
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28 tháng 9 2016
\(\frac{3}{2}x^2+y^2+z^2+yz=1\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)
Suy ra : \(A^2\le2\Rightarrow A\le\sqrt{2}\)
Vậy Max A = \(\sqrt{2}\) khi \(\hept{\begin{cases}x=y\\x=z\\x+y+z=\sqrt{2}\end{cases}\Leftrightarrow}x=y=z=\frac{\sqrt{2}}{3}\)
NH
24 tháng 2 2020
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2019}{y}=\frac{x+y-2020}{z}=\frac{y+z+1+x+z+2019+x+y-2020}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow2=\frac{1}{x+y+z}\)\(\Rightarrow x+y+z=\frac{1}{2}\)
Ta có:
+) \(\frac{y+z+1}{x}=2\)\(\Rightarrow y+z+1=2x\)\(\Rightarrow x+y+z+1=3x\)\(\Rightarrow\frac{1}{2}+1=3x\)\(\Rightarrow3x=\frac{3}{2}\)\(\Rightarrow x=\frac{1}{2}\)
+) \(\frac{x+z+2019}{y}=2\)\(\Rightarrow x+z+2019=2y\)\(\Rightarrow x+y+z+2019=3y\)\(\Rightarrow\frac{1}{2}+2019=3y\)\(\Rightarrow3y=\frac{4039}{2}\)\(\Rightarrow y=\frac{4039}{6}\)
+) \(\frac{x+y-2020}{z}=2\)\(\Rightarrow x+y-2020=2z\)\(\Rightarrow x+y+z-2020=3z\)\(\Rightarrow\frac{1}{2}-2020=3z\)\(\Rightarrow3z=\frac{-4039}{2}\)\(\Rightarrow z=\frac{-4039}{6}\)
Lại có: \(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{4039}{6}\right)^{2017}+\left(\frac{-4039}{6}\right)^{2017}=4032+\left(\frac{4039}{6}\right)^{2017}-\left(\frac{4039}{6}\right)^{2017}=4032\)
HP
12 tháng 8 2016
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.....................
Y
9 tháng 6 2019
+ Theo bđt cauchy :
\(\frac{1}{x^2+x}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}\cdot\frac{x}{2}\cdot\frac{x+1}{4}}=\frac{3}{2}\)
Dấu "=" \(\Leftrightarrow\frac{1}{x\left(x+1\right)}=\frac{x}{2}=\frac{x+1}{4}\Leftrightarrow x=1\)
+ Tương tự :
\(\frac{1}{y^2+y}+\frac{y}{2}+\frac{y+1}{4}\ge\frac{3}{2}\) Dấu "=" <=> y = 1
\(\frac{1}{z^2+z}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}\) Dấu "=" <=> z = 1
Do đó : \(P+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)
\(\Rightarrow P+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\) \(\Rightarrow P\ge\frac{3}{2}\)
Dấu "=" <=> x = y = z = 1
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 \(S=-\frac{1}{7}\)