Violympic toán 9
Các câu hỏi tương tự
Chứng minh rằng nếu \(\text{ax}^3=by^3=cz^3\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) thì
\(\sqrt[3]{\text{ax}^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
Cho \(ax^3=by^3=cz^3;\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1.\)C/m \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
a) cmr (ax+by+cz)\(^2\)≤\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
b) cho a,b,c >0 tm \(a^2+b^2+c^2=1\)
cmr :\(\frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\le\frac{a}{2\left(a+b+c\right)}\)
cho 3 so thuc x,y,z khac khong va thoa man hai dieu kien \(ax^3=by^3=cz^3\) va \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
chung minh rang : \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
2 tháng 3 2020 lúc 23:45
1. a) left{{}begin{matrix}x,y,z0xyz1end{matrix}right.. Tìm max Pfrac{1}{sqrt{x^5-x^2+3xy+6}}+frac{1}{sqrt{y^5-y^2+3yz+6}}+frac{1}{sqrt{z^5-z^2+zx+6}}
b) left{{}begin{matrix}x,y,z0xyz8end{matrix}right.. Min Pfrac{x^2}{sqrt{left(1+x^3right)left(1+y^3right)}}+frac{y^2}{sqrt{left(1+y^3right)left(1+z^3right)}}+frac{z^2}{sqrt{left(1+z^3right)left(1+x^3right)}}
c) x,y,z0. Min Psqrt{frac{x^3}{x^3+left(y+zright)^3}}+sqrt{frac{y^3}{y^3+left(z+xright)^3}}+sqrt{frac{z^3}{z^3+left(x+yright)^3}}
d) a,b,c0;...
Đọc tiếp
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
1 ) Cho a,b,c >0 và abc= 1.CMR:
\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
2 ) Cho x,y,z > 0 và x+y+z=3
CMR : \(\frac{x}{x+\sqrt{3x+yz}}+\frac{y}{y+\sqrt{3y+zx}}+\frac{z}{z+\sqrt{3z+xy}}\le1\)
1.
a/ cho 6 số dương a,b,c,x,y,z thỏa mãn : ax+by+czxyz. cmr: x+y+zsqrt{a+b}+sqrt{b+c}+sqrt{c+a}
b/ cm: sqrt{frac{a}{b+c}}+sqrt{frac{b}{a+c}}+sqrt{frac{c}{b+c}}2 với a,b,c 0
2.
a/ cho left(x+sqrt{x^2+2013}right).left(y+sqrt{y^2+2013}right)2013
b/ cho a,b là các số tự nhiên .cmr : 5a^2+15ab-b^2⋮49Leftrightarrow3a+b⋮7
Đọc tiếp
1.
a/ cho 6 số dương a,b,c,x,y,z thỏa mãn : ax+by+cz=xyz. cmr: \(x+y+z>\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
b/ cm: \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+c}}>2\) với a,b,c >0
2.
a/ cho \(\left(x+\sqrt{x^2+2013}\right).\left(y+\sqrt{y^2+2013}\right)=2013\)
b/ cho a,b là các số tự nhiên .cmr : \(5a^2+15ab-b^2⋮49\Leftrightarrow3a+b⋮7\)
Cho các số dương x,y,z thỏa mãn điều kiện xy+yz+zx=1
CMR: \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1=z^2}}\le\frac{3}{2}\)
Cho \(x\ge3,y\ge2,z\ge1.CMR\)
\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\le\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)