Những câu hỏi liên quan
Cho x, y, z > 0 thoả mãn: \(xy+yz+zx=6xyz\). Tìm giá trị nhỏ nhất của \(T=x\sqrt{\dfrac{x}{1+x^3}}+y\sqrt{\dfrac{y}{1+y^3}}+z\sqrt{\dfrac{z}{1+z^3}}\)
\(xy+xz+yz=6xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y}=b\\\frac{1}{z}=c\end{matrix}\right.\) \(\Rightarrow a+b+c=6\)
\(T=\sum x\sqrt{\frac{x}{1+x^3}}=\sum\sqrt{\frac{x^3}{1+x^3}}=\sum\sqrt{\frac{1}{1+\frac{1}{x^3}}}=\sum\frac{1}{\sqrt{1+a^3}}=\sum\frac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\Rightarrow T\ge\sum\frac{2}{a+1+a^2-a+1}=\sum\frac{2}{a^2+2}\)
Ta có đánh giá: \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\) với mọi \(0< a< 6\)
Thật vậy, \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\Leftrightarrow18-\left(a^2+2\right)\left(7-2a\right)\ge0\)
\(\Leftrightarrow2a^3-7a^2+4a+4\ge0\)
\(\Leftrightarrow\left(a-2\right)^2\left(2a+1\right)\ge0\) luôn đúng với mọi \(0< a< 6\)
Tương tự ta có: \(\frac{2}{b^2+2}\ge\frac{7-2b}{9}\) ; \(\frac{2}{c^2+2}\ge\frac{7-2c}{9}\)
\(\Rightarrow T\ge\frac{21-2\left(a+b+c\right)}{9}=\frac{21-12}{9}=1\)
\(\Rightarrow T_{min}=1\) khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)
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cm P=1/x+y+z * 1/xy+yz+zx * (1/y+1/x+1/z)*(1/xy+1/yz+1/zx) luôn dương với mọi giá trị của x,y,z
chứng minh rằng: (x-y)/(1+xy) + (y-z)/(1+yz) +(z-x)/(1+zx) = (x-y)(y-z)(z-x)/(1+xy)(1+yz)(1+zx)
Ta có:
\(\dfrac{x-y}{1+xy}\)+\(\dfrac{y-z}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) = \(\dfrac{x-y}{1+xy}\)+\(\dfrac{-\left(x-y\right)-\left(z-x\right)}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)
=\(\dfrac{x-y}{1+xy}\)\(-\dfrac{x-y}{1+yz}\) \(-\dfrac{z-x}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)
= \(\left(x-y\right)\)\(\left(\dfrac{\left(1+yz\right)-\left(1+xy\right)}{\left(1+yz\right)\left(1+xy\right)}\right)\)+(\(z-x\))\(\left(\dfrac{\left(1+yz\right)-\left(1+zx\right)}{\left(1+yz\right)\left(1+zx\right)}\right)\)
=\(\left(x-y\right)\)\(\dfrac{y\left(z-x\right)}{\left(1+yz\right)\left(1+xy\right)}\)+(\(z-x\))\(\dfrac{-z\left(x-y\right)}{\left(1+yz\right)\left(1+zx\right)}\)
=\(\left(\dfrac{\left(x-y\right)\left(z-x\right)}{1+yz}\right)\)\(\left(\dfrac{y\left(1+xz\right)-z\left(1+xy\right)}{\left(1+xz\right)\left(1+xy\right)}\right)\)
=đpcm
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chứng minh A=(xy+zx+1)/(xy+x+y+1)+(yz+zy+1)/(yz+y+z+1)+(zx+zx+1)/(zx+x+z+1) không thuộc x, y, z
làm nhanh giùm mình nha ! đang cần gấp <:)
Cho x,y,z>0 thỏa mãn xy+yz+zx=1. Chứng minh \(\frac{x}{x^2-yz+3}+\frac{y}{y^2-zx+3}+\frac{z}{z^2-xy+3}\ge\frac{1}{x+y+z}\)
cho x, y, z khác 1 chứng minh giá trị sau không phụ thuộc vào biến x, y, z.( xy+2x+1/xy+x+y+1)+(yz+2y+1/yz+y+z+1)+(zx+2z+1/zx+z+x+1)
Sửa lại đề là x;y;z khác -1.
\(A=\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+z+x+1}=\)
\(A=\frac{x\left(y+1\right)+x+1}{x\left(y+1\right)+y+1}+\frac{y\left(z+1\right)+y+1}{y\left(z+1\right)+z+1}+\frac{z\left(x+1\right)+z+1}{z\left(x+1\right)+x+1}=\)
\(A=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\frac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\frac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}=\)vì x;y;z khác -1 nên:
\(A=\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}=\)
\(A=\frac{x}{x+1}+\frac{1}{x+1}+\frac{y}{y+1}+\frac{1}{y+1}+\frac{z}{z+1}+\frac{1}{z+1}=\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}=1+1+1=3\)
A = 3 với mọi x;y;z khác -1 nên A không phụ thuộc vào x;y;z. đpcm
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1.Giải hệ pt1)hept{begin{cases}frac{1}{x}+frac{1}{y}+frac{1}{z}3xy+yz+zx3frac{1}{1+x+xy}+frac{1}{1+y+yz}+frac{1}{1+z+zx}xend{cases}}2)hept{begin{cases}xy+yz+zx3left(x+yright)left(y+zright)sqrt{3}zleft(1+y^2right)left(y+zright)left(z+xright)sqrt{3}xleft(1+z^2right)end{cases}}3)hept{begin{cases}xy+yz+zx31+x^2left(y+zright)+xyz4y1+y^2left(z+xright)+xyz4zend{cases}}
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1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Cho x; y; z >0, thoả mãn: 1/xy+ 1/yz+1/zx =1
Q= x/√yz × (x^2 +1)+ y/√zx × (y^2 +1) + z/√xy × ( z^2 +1)
cho 3 số dương 0<x<y<z<1.CM X/YZ+1 + Y/ZX+1 + Z/XY+1<2