Áp dụng bất đẳng thức cô si ta có :
\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3+\left(\sqrt{z}\right)^3\ge3\sqrt[3]{\left(\sqrt{xyz}\right)^3}=3\sqrt{xyz}\)Dấu "=" xảy ra khi :\(\sqrt{x}=\sqrt{y}=\sqrt{z}\)
Do đó :\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Vậy A=8
tìm x,y,z biết
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)
Tìm x,y,z
\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) (Với x+y+z=3)
cho x,y,z,a là các số dương;\(a^2=b+4028và\left\{{}\begin{matrix}x+y+z=a\\x^2+y^2+z^2=b\end{matrix}\right.\).tính:
S=\(x\sqrt{\dfrac{\left(2014+y^2\right)\left(2014+z^2\right)}{2014+x^2}}\)+\(y\sqrt{\dfrac{\left(2014+z^2\right)\left(2014+x^2\right)}{2014+y^2}}\)+z\(\sqrt{\dfrac{\left(2014+x^2\right)\left(2014+y^2\right)}{2014+z^2}}\)
Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
tìm 3 số thực dương x;y;z thỏa mãn \(\dfrac{2}{\sqrt{x}+2\sqrt{y}+3\sqrt{z}}-\dfrac{1}{2\sqrt{xy}+6\sqrt{yz}+3\sqrt{zx}}=\dfrac{1}{3}\)
Tìm x,y,z,
\(\sqrt{x}\)+\(\sqrt{y-1}\)+\(\sqrt{z-2}\)=\(\dfrac{1}{2}\left(x+y+z\right)\)
Tìm x, y, z biết \(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\frac{1}{2}\left(x+y+z\right)\)
Giải phương trình:
1) \(x^2-4x-2\sqrt{2x-5}+5=0\)
2)\(x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
3)\(\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
4)\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
