\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) . Tìm Min \(\sqrt{\frac{2x^{3}+3y^{2}}{x+4y}}+\sqrt{\frac{2y^{3}+3z^{2}}{y+4z}}+\sqrt{\frac{2z^{3}+3x^{2}}{z+4x}}\)
1) Rút gọn : \(\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
2) CHo \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\). CMR \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
Bài 1:
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giải hệ phương trình :
a) \(\hept{\begin{cases}x^2+y^2=1\\x^9+y^9=1\end{cases}}\)
b)\(\hept{\begin{cases}\sqrt{x}+\sqrt{y}+\sqrt{z}=2014\\\frac{1}{3x+2y}+\frac{1}{3y+2z}+\frac{1}{3z+2x}=\frac{1}{x+2y+3z}+\frac{1}{y+2x+3x}+\frac{1}{z+2x+3y}\end{cases}}\)
google xin tài trợ chương trình
có google thôi anh
Rút gọn: M = \(\frac{5x^5+4x^4+3x^3+2}{4x^4+3x^3+2x^2+z}+\frac{4y^4+3y^3+2y^2+y}{5y^5+4y^4+3y^3+2}+\frac{5y^5+4z^4+3z^3+2}{4z^4+3z^3+2z^2+z}\)
Cho x,y,z>0 :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}\)
Tìm min P=\(\frac{\sqrt{2x^2+y^2}}{xy}+\frac{\sqrt{2y^2+z^2}}{yz}+\frac{\sqrt{2z^2+x^2}}{zx}\)
gọi P là cái 1/x+1/y+1/z nha
1) (1/x+1/y+1/z)^2 = 1/x^2 + 1/y^2 + 1/z^2 + 2/(xy) + 2/(yz) + 2/(zx)
---> 3 = P + 2(x+y+z)/(xyz) = P + 2 ---> P = 1
a)Cho \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)và 3x-2y+z=40.Tìm x,y,z
b)Tìm x,y biết \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
Giúp mik với!help me~~~
a, cho 2 số dương x,y thỏa mãn x+y=1
tìm min của \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)\)
b, cho x,y,z là các số dương thỏa mãn : \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\)
cmr : \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
Cho x,y,z>0 và x+y+z=3. Tìm Min A = \(\frac{z}{\sqrt{x^2+5xy+4y^2}}+\frac{x}{\sqrt{y^2+5yz+4z^2}}+\frac{y}{\sqrt{z^2+5zx+4x^2}}\)
Cho \(\hept{\begin{cases}x,y,z>0\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\end{cases}}\)Tìm min A = \(\frac{\sqrt{x^2+2y^2}}{xy}+\frac{\sqrt{y^2+2z^2}}{yz}+\frac{\sqrt{z^2+2x^2}}{zx}\)
Ta có \(\frac{\sqrt{x^2+2y^2}}{xy}=\sqrt{\frac{1}{y^2}+\frac{2}{x^2}}\)
Áp dụng BĐT Buniacoxki ta có
\(\sqrt{\left(\frac{1}{y^2}+\frac{2}{x^2}\right)\left(1+2\right)}\ge\sqrt{\left(\frac{1}{y}+\frac{2}{x}\right)^2}=\frac{1}{y}+\frac{2}{x}\)
=> \(\sqrt{3}A\ge3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3\)
=> \(A\ge\sqrt{3}\)
\(MinA=\sqrt{3}\)khi x=y=z=3
cho cac si thuc duong x,y,z thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
tìm Max của P=\(\frac{1}{\sqrt{2x^2+y^2+3}}+\frac{1}{\sqrt{2y^2+z^2+3}}+\frac{1}{\sqrt{2z^2+x^2+3}}\)