Violympic toán 9
Các câu hỏi tương tự
a)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x-y=7\\x^3+y^3=133\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2-5x+y=0\\x-\sqrt{y}+1=0\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\zy+yz+zx=27\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}x\left(yz+1\right)=\frac{7}{3}z\\y\left(xz+1\right)=8x\\z\left(xy+1\right)=\frac{9}{2}y\end{matrix}\right.\)
giải hệ phương trình : \(\left\{{}\begin{matrix}x+y+z=9\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)
1. \(\left\{{}\begin{matrix}x^2-yz=a\\y^2+xz=b\\z^2+xy=c\end{matrix}\right.\) Tính \(\frac{ax-by-cz}{x-y+z}\)theo a,b,c
2. \(x^2+y^2+\frac{9}{2}z^2=5\). Tìm maxA \(=xy+yz+zx\)
25 tháng 4 2020 lúc 12:03
2. a) left{{}begin{matrix}x,y,z1x+y+zxyzend{matrix}right. Tìm min Pfrac{x-1}{y^2}+frac{y-1}{z^2}+frac{z-1}{x^2}
b) a,b,c0.Cmr:left(frac{a}{b}+frac{b}{c}+frac{c}{a}right)^2geleft(a+b+cright)left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)
c) left{{}begin{matrix}x,y,zge0x^2+y^2+z^22end{matrix}right. Tìm max Pfrac{x^2}{x^2+yz+x+1}+frac{y+z}{x+y+z+1}-frac{1+yz}{9}
d) left{{}begin{matrix}a,b,c0a+b+c3end{matrix}right.. Cmr: frac{a}{ab+3c}+frac{b}{bc+3a}+frac{c}{ca+3b}gefrac{3}{4}
Đọc tiếp
2. a) \(\left\{{}\begin{matrix}x,y,z>1\\x+y+z=xyz\end{matrix}\right.\) Tìm min \(P=\frac{x-1}{y^2}+\frac{y-1}{z^2}+\frac{z-1}{x^2}\)
b) \(a,b,c>0.Cmr:\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
c) \(\left\{{}\begin{matrix}x,y,z\ge0\\x^2+y^2+z^2=2\end{matrix}\right.\) Tìm max \(P=\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}\)
d) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{ab+3c}+\frac{b}{bc+3a}+\frac{c}{ca+3b}\ge\frac{3}{4}\)
Giải hpt: \(\left\{{}\begin{matrix}x+xy+y=9\\y+yz+z=4\\z+zx+x=1\end{matrix}\right.\)
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 : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
Giải hệ: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}-\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{z}}=\frac{8}{3}\\x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{118}{9}\\x\sqrt{x}+y\sqrt{y}+z\sqrt{z}-\frac{1}{x\sqrt{x}}-\frac{1}{y\sqrt{y}}-\frac{1}{z\sqrt{z}}=\frac{728}{27}\end{matrix}\right.\)