Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{3}{2}x^2+y^2+z^2+yz=1\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)
Suy ra : \(A^2\le2\Rightarrow A\le\sqrt{2}\)
Vậy Max A = \(\sqrt{2}\) khi \(\hept{\begin{cases}x=y\\x=z\\x+y+z=\sqrt{2}\end{cases}\Leftrightarrow}x=y=z=\frac{\sqrt{2}}{3}\)
+ 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
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=\(^{x^2+y^2}\)
Trả lời
Từ giả thiết x+y+z=xyz <=> 1/xy + 1/yz + 1/zx = 1
Khi đó: x/1+x2 = \(\frac{1}{\frac{x}{\left(\frac{1}{z}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}}\)\(=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có:\(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra VT=\(\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
ĐPCM
Ta có:\(\frac{x}{1+x^2}=\frac{xyz}{yz+x^2yz}=\frac{xyz}{yz+x\left(xyz\right)}=\frac{xyz}{yz+x\left(x+y+z\right)}=\frac{xyz}{yz+x^2+xy+xz}=\frac{xyz}{y\left(x+z\right)+x\left(x+z\right)}\)
\(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}\)
Chứng minh tương tự : \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(y+z\right)\left(y+x\right)}\)
\(\frac{3z}{1+z^2}=\frac{3xyz}{\left(x+z\right)\left(x+y\right)}\)
Khi đó VT \(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}+\frac{2xyz}{\left(y+z\right)\left(y+x\right)}+\frac{3xyz}{\left(x+z\right)\left(z+y\right)}\)
\(=\frac{xyz\left[y+z+2\left(z+x\right)+3\left(x+y\right)\right]}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(đpcm\right)\)
( mình đang vội nên làm hơi tắt mong bạn thông cảm )
Ta luôn có:
\(xy+yz+zx\le x^2+y^2+z^2\)\(=3\); dấu "=" xảy ra ⇔\(x=y=z\)
\(x\le\frac{x^2+1}{2}\); dấu "=" xảy ra ⇔ \(x=1\)
\(y\le\frac{y^2+1}{2}\); dấu "=" xảy ra ⇔ \(y=1\)
\(z\le\frac{z^2+1}{2}\); dấu "=" xảy ra ⇔ \(z=1\)
Suy ra: \(x+y+z\le\frac{x^2+y^2+z^2+3}{2}=\frac{6}{2}=3\)
Do đó: \(P_{max}=xy+yz+zx+\frac{5}{x+y+z}\le3+\frac{5}{3}=\frac{14}{3}\)
Dấu "=" xảy ra ⇔ 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 \(S=-\frac{1}{7}\)