Cho A=\(\frac{6x^2z^2}{y^2}+\frac{8y^2z^2}{x^2}+\frac{10x^2y^2}{z^2}\)biết 2xy +yz=3. Tìm GTNN của A
cho x, y, z khác 0 thoả mãn : 2xy+yz=3
tìm GTLN của A= 6x^2z^2/y^2 + 8y^2z^2/x^2 + 10x^2y^2/z^2
Tìm các số x, y, z biết: \(\frac{xy}{2y+4x}=\frac{yz}{4x+6y}=\frac{zx}{6x+2z}=\frac{x^2+y^2+z^2}{2^2+4^2+6^2}\)
Nếu một trong các số x,y,z bằng không thì dễ thấy các số còn lại cũng bằng 0
Suy ra x;y;z khác 0
Đặt \(2=a;4=b;6=c\) khi đó ta có:
\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}\)
\(\Rightarrow\frac{xyz}{ayz+bxz}=\frac{xyz}{bxz+xcy}=\frac{xyz}{cyx+ayz}\)
Mà \(x;y;z\ne0\) suy ra:
\(ayz+bxz=bxz+xcy=cxy+ayz\)
\(\Rightarrow az=cx;bx=ay\)
\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Rightarrow x=ak;y=bk;z=ck\)
Khi đó:\(\frac{xy}{ay+bx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{ak\cdot bk}{abk+abk}=\frac{a^2k^2+b^2k^2+c^2k^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{k}{2}=k^2\)
\(\Rightarrow k=\frac{1}{2}\)
\(\Rightarrow x=\frac{a}{2};y=\frac{b}{2};z=\frac{c}{2}\)
Thay số vào,ta được:
\(x=1;y=2;z=3\)
Tìm các số x, y, z biết: \(\frac{xy}{2y+4x}=\frac{yz}{4x+6y}=\frac{zx}{6x+2z}=\frac{x^{2}+y^{2}+z^{2}}{2^{2}+4^{2}+6^{2}}\)
\(\frac{xy}{2y+4x}=\frac{yz}{4z+6y}=\frac{xz}{6x+2z}\)(4z chứ 4x là sai đề rồi bạn)
\(\Leftrightarrow\frac{x}{2}+\frac{y}{4}=\frac{y}{4}+\frac{z}{6}=\frac{z}{6}+\frac{x}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)tự làm tiếp :))
Tìm các số x, y, z biết: \(\frac{xy}{2y+4x}=\frac{yz}{4x+6y}=\frac{zx}{6x+2z}=\frac{x^{2}+y^{2}+z^{2}}{2^{2}+4^{2}+6^{2}}\)
sorry sai đề :v
Sửa \(\frac{xy}{2y+4x}+\frac{yz}{4z+6y}=\frac{zx}{6x+2z}=\frac{x^2+y^2+z^2}{2^2+4^2+6^2}\)
Ta có :
\(\frac{xy}{2y+4x}=\frac{yz}{4z+6y}=\frac{zx}{6x+2z}=\frac{x^2+y^2+z^2}{2^2+4^2+6^2}\)
\(\Leftrightarrow\frac{xyz}{2yz+4xz}=\frac{xyz}{4xz+6xy}=\frac{xyz}{6xy+2yz}\)
\(\Rightarrow2yz+4xz=4xz+6xy=6xy+2yz\)
\(\Rightarrow\hept{\begin{cases}2yz=6xy\\4xz=2yz\end{cases}}\Leftrightarrow\hept{\begin{cases}z=3x\\y=2x\end{cases}}\)
\(\rightarrow x:y:z=1:2:3\frac{xy}{2y+4x}\) \(=\frac{yz}{4z+6y}=\frac{zx}{6x+2z}=\frac{2x^2}{4y+4x}=\frac{x}{4}.\frac{x^2+y^2+z^2}{2^2+4^2+6^2}=\frac{14x^2}{56}=\frac{x^2}{4}\rightarrow\frac{x^2}{4}=\frac{x}{4}\)
\(\Rightarrow\frac{x^2-x}{4}=0\Leftrightarrow x-1=0\left(x\ne0\right)\)
\(\Rightarrow x=1\rightarrow x=1;y=2;z=3\)
Làm thử thôi sai thì thôi nha !
Cho xy+yz+zx=2xyz ; x,y,z>0 Tìm max \(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)
Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:
\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)
\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)
Cho x,y,z >0 thỏa mãn: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}\)
Tìm GTNN của: \(P=\frac{\sqrt{2x^2+y^2}}{xy}+\frac{\sqrt{2y^2+z^2}}{yz}+\frac{\sqrt{2z^2+x^2}}{zx}\)
Áp dụng bđt bu nhi a cốp xki :
\(\left(2x^2+y^2\right)\left(\left(\sqrt{2}\right)^2+\left(1\right)^2\right)\ge\left(\sqrt{2}.\sqrt{2}x+y.1\right)^2=\left(2x+y\right)^2\)
=> \(\sqrt{2x^2+y^2}\ge\frac{1}{\sqrt{3}}\left(2x+y\right)\) => \(\frac{\sqrt{2x^2+y^2}}{xy}\ge\frac{1}{\sqrt{3}}\cdot\frac{2x+y}{xy}=\frac{1}{\sqrt{3}}\left(\frac{2}{y}+\frac{1}{x}\right)\)
CM tương tự với hai cái còn lại
=> \(P\ge\frac{1}{\sqrt{3}}\left(\frac{3}{x}+\frac{3}{y}+\frac{3}{z}\right)=\frac{1}{\sqrt{3}}\cdot3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{\sqrt{3}}\cdot3\cdot\sqrt{3}=3\)
Dấu '' = '' xảy ra khi x = y =z = căn 3
Cho x,y,z thỏa mãn x+y+z=\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\). Chứng minh rằng
\(\frac{1}{\left(2xy+yz+zx\right)^2}+\frac{1}{\left(2yz+zx+xy\right)^2}+\frac{1}{\left(2xz+xy+yz\right)^2}\le\frac{3}{16x^2y^2z^2}\)
Cho xy+yz+xz=2xyz (x,y,z>0). Tìm Max P= \(\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2z^2x^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)