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ta có \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)
ta có :\(\sqrt{4x+5}=\frac{\sqrt{9\left(4x+5\right)}}{3}\le\frac{9+4x+5}{2\times3}=\frac{2x+7}{3}\)
tương tự ta sẽ có ; \(A\le\frac{2x+7}{3}+\frac{2y+7}{3}+\frac{2z+7}{3}=\frac{2}{3}\left(x+y+z\right)+7\le\frac{2}{3}\times3+7=9\)
Vậy GTLN của A=9
dấu bằng xảy ra khi x= y= z =1
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2\left(xy+yz+zx\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3.3=9\)
\(\Rightarrow x+y+z\le3\).
\(A=\sqrt{4x+5}+\sqrt{4y+5}+\sqrt{4z+5}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4x+5+4y+5+4z+5\right)}\)
\(=\sqrt{3\left[4\left(x+y+z\right)+15\right]}=9\)
Dấu \(=\)khi \(x=y=z=1\).
1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2
= 4/9 .y.y.y . (3/2-3/2.y)^2
=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)
<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5
=4/9 . 243/3125
=108/3125
Đến đó tự giải
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
Bài 1: ĐK:....
Cộng theo vế 3 pt trên ta có
\(2x+2y+2z=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\)
\(\Leftrightarrow4x+4y+4z-2\sqrt{4x-1}-2\sqrt{4y-1}-2\sqrt{4z-1}=0\)
\(\Leftrightarrow\left(4x-1-2\sqrt{4x-1}+1\right)+\left(4y-1-2\sqrt{4y-1}+1\right)+\left(4z-1-2\sqrt{4z-1}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)
Xảy ra khi \(\hept{\begin{cases}\sqrt{4x-1}=1\\\sqrt{4y-1}=1\\\sqrt{4z-1}=1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}4x-1=1\\4y-1=1\\4z-1=1\end{cases}}\)\(\Rightarrow x=y=z=\frac{1}{2}\)
Ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Áp dụng BĐT Cosi ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)
Cmtt:
\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)
\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)
Bài 32:
a) P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) Có: \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(y+x\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x-2y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=2y\end{cases}}}\)
Thay x=-y ta có: Q=\(\frac{-y-y}{-y+y}\)=\(\frac{-2y}{0}\)(loại )
Thay x=2y ta có : Q=\(\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)