Bài 1:tìm x;y sao
x.y-2y+x=8
Bài 2: tìm x;y;z thuộc
a)xy=-35;yz=7;xyz=35
b)xy=-6;yz=-15;xyz=-30;xyzt=120
Cho các số thực dương x,y,z thõa mãn \(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}=\sqrt{xyz}\)
Tìm giá trị nhỏ nhất của biểu thức
P=\(\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Cho xy+yz+xz=2xyz (x,y,z>0). Tìm Max P= \(\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2z^2x^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
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 thỏa mãn xyz=1. tìm GTNN của \(T=\dfrac{xy}{z^2x+z^2y}+\dfrac{yz}{x^2y+x^2z}+\dfrac{zx}{y^2x+y^2z}\)
\(T=\dfrac{\left(xy\right)^2}{zx+zy}+\dfrac{\left(yz\right)^2}{xy+xz}+\dfrac{\left(zx\right)^2}{yx+yz}\ge\dfrac{xy+yz+zx}{2}\ge\dfrac{3}{2}\sqrt[3]{\left(xyz\right)^2}=\dfrac{3}{2}\)
\(\dfrac{xyz-xy-yz-zx+x+y+z-1}{xyz+xy+yz-zx-x+y-z-1}\) với x = 5001;y=5002;z=5003
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
Cho các số thực dương x, y, z thõa mãn \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\). tìm giá trị nhỏ nhất của biểu thức P=\(\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)
Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).
\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).
Ta có:
\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)
\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).
Ta có:
\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).
\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).
\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).
\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).
\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).
Chứng minh tương tự, ta được:
\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).
Chứng minh tương tự, ta được:
\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).
\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).
\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)
\(\left(4\right)\).
Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).
\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)
(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).
\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(P\ge\frac{\sqrt{5}}{3}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).
Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).
Tìm x,y,z,t biết xyzt=1
Tính P=1/(1+x+xy+xyz)+1/(1+y+yz+yzt)+1/(1+z+zt+ztx)+1/(1+t+tx+txy)
mọi người giúp mik với sắp đi hok r mà vẫn chưa xong hết bài =(((
Answer:
\(P=\frac{1}{1+x+xy+xyz}+\frac{1}{1+y+yz+yzt}+\frac{1}{1+z+zt+ztx}+\frac{1}{1+t+tx+txy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+xyzt}+\frac{xy}{xy+xyz+xyzt+xyzt.x}+\frac{xyz}{xyz+xyzt+xyzt.x+xyzt.xy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+1}+\frac{xy}{xy+xyz+1+x}+\frac{xyz}{xyz+1+x+xy}\)
\(=\frac{1+x+xy+xyz}{1+x+xy+xyz}\)
\(=1\)
Cho xyzt=1 tính tổng (x/xyz+xy+x+1)+(y/yzt+yz+yt+1)+(z/zxt+zt+z+1)+(t/xyt+t+t+1)
1. Tìm các số nguyên x,y,z biết:
A) xy=-35, yz=7, xyz=35
B) x+y=-1,x+z=6,z+y=1
a: Sửa đề: xy=35; yz=5
\(\left(xy\cdot yz\cdot xz\right)=\left(35\cdot7\cdot5\right)^2=35^2=35\)
=>z=1; x=5; y=7
b: \(x+y+z=\dfrac{-1+6+1}{2}=3\)
=>z=3+1=4; y=3-6=-3; x=3-1=2