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Ta có:\(10=2xyz\)
=> \(P=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+10}+\frac{10z}{10z+yz+10}\)
\(=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+2xyz}+\frac{2xyz^2}{2xyz^2+yz+2xyz}\)
\(=\frac{1}{2x+2xz+1}+\frac{2x}{1+2x+2xz}+\frac{2xz}{2xz+1+2x}\)
\(=1\)
Vậy P=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 \(S=-\frac{1}{7}\)
+ 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
a, Xét : 196 = 14^2 = (a^2+b^2+c^2) = a^4+b^4+c^4+2.(a^2b^2+b^2c^2+c^2a^2)
<=> a^4+b^4+c^4 = 196 - 2.(a^2b^2+b^2c^2+c^2a^2)
Xét : 0 = (a+b+c)^2 = a^2+b^2+c^2+2.(ab+bc+ca)
Mà a^2+b^2+c^2 = 14
<=> 2.(ab+bc+ca) = -14
<=> ab+bc+ca = -7
<=> a^2b^2+b^2c^2+c^2a^2+2abc.(a+b+c) = 49
Lại có : a+b+c = 0
<=> a^2b^2+b^2c^2+c^2a^2 = 49
<=> A = a^4+b^4+c^4 = 196 - 2.49 = 98
Tk mk nha
b) \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\)\(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow\)\(x^2=y^2=z^2=0\)
\(\Leftrightarrow\)\(x=y=z=0\)
Vậy \(D=0\)
theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
Từ \(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{x+z}\Rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{x+z}{xz}\)
\(\Rightarrow\frac{x}{xy}+\frac{y}{xy}=\frac{y}{yz}+\frac{z}{yz}=\frac{x}{xz}+\frac{z}{xz}\)
\(\Rightarrow\frac{1}{y}+\frac{1}{x}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\).Khi đó
\(P=\frac{20xy+4yz+2013xz}{x^2+y^2+z^2}=\frac{20x^2+4x^2+2013x^2}{x^2+x^2+x^2}=\frac{2037x^2}{3x^2}=679\)
cho x,y>0 thỏa mãn \(^{x^2+y^2-xy=8}\)
tìm GTNN và GTNN của biểu thức M=\(^{x^2+y^2}\)
\(\Leftrightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=...\\y=...\\z=...\end{matrix}\right.\)