tìm x y z thuộc Z
biết x^2+4=4y
y^2+4=4z
z^2+4=4x
Cho x,y,z>0 sao cho x+y+z=5. Tìm gtnn của A=\(\dfrac{4x}{y^2+4}+\dfrac{4y}{z^2+4}+\dfrac{4z}{x^2+4}\)
Cho x,y,z>0 sao cho x+y+z=5. Tìm gtnn của A=\(\dfrac{4x}{y^2+4}+\dfrac{4y}{z^2+4}+\dfrac{4z}{x^2+4}\)
Cho 1/x+1/y+1/z=0.CMR:(x^2y^2+y^2z^2+z^2x^2)^2=2(x^4y^4+y^4z^4+z^4x^4)
Cho 0<x,y,z<\(\dfrac{\sqrt{3}}{2}\) thỏa mãn xy+yz+zx=\(\dfrac{3}{4}\)
Tìm Min Q=\(\dfrac{4x^2}{x\left(32-4x^2\right)}+\dfrac{4y^2}{y\left(32-4y^2\right)}+\dfrac{4z^2}{z\left(32-4z^2\right)}\)
Cho 0<x,y,z<\(\dfrac{\sqrt{3}}{2}\) thỏa mãn xy+yz+zx=\(\dfrac{3}{4}\)
Tìm Min \(Q=\dfrac{4x^2}{x\left(3-4x^2\right)}+\dfrac{4y^2}{y\left(3-4y^2\right)}+\dfrac{4z^2}{z\left(3-4z^2\right)}\)
Ta chứng minh BĐT sau:
Ta có: \(x\left(3-4x^2\right)=-4x^3+3x-1+1=1-\left(x+1\right)\left(2x-1\right)^2\le1\)
\(\Rightarrow\dfrac{4x^2}{x\left(3-4x^2\right)}\ge\dfrac{4x^2}{1}=4x^2\)
Tương tự và cộng lại:
\(Q\ge4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)=3\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{2}\)
tìm x y z
biết x^2+4=4y
y^2+4=4z
z^2+4=4x
Tìm x; y; z:
Cộng vế theo vế ta có:
\(x^2+y^2+z^2+4+4+4=4y+4z+4x\)
<=> \(\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+\left(z^2-4z+4\right)=0\)
<=> \(\left(x-2\right)^2+\left(y-2\right)^2+\left(z-2\right)^2=0\)
<=> x = y = z = 2
Cho x,y,z thỏa mãn x^2+4y+4=0 và y^2+4z+4=0 và z^2+4x+4=0. Tính x^10+y^10+z^10
Ta có:\(x^2+4y+4=0;y^2+4z+4=0;z^2+4x+4=0\)
\(\Leftrightarrow\left(x^2+4y+4\right)+\left(y^2+4z+4\right)+\left(z^2+4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4+y^2+4y+4+z^2+4z+4=0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2=0\)
Mà\(\left(x+2\right)^2\ge0;\left(y+2\right)^2\ge0;\left(z+2\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)
Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y+2=0\\z+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-2\\z=-2\end{cases}\Leftrightarrow}x=y=z=-2}\)
Vậy\(x^{10}+y^{10}+z^{10}=x^{10}+x^{10}+x^{10}\)
\(=3\cdot x^{10}=3\cdot\left(-2\right)^{10}=3\cdot1024=3072\)
1, x : y : z = 2 : 3 : 4 và x + y + z = 18
2, \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}\) và 4x - 3y - 2z = 81
3, \(\dfrac{x}{3}=\dfrac{y}{2};\) 4y = 3z và x + y +z = 46
4, 5x = 3y; \(\dfrac{y}{z}=\dfrac{3}{2}\) và 2x + 3y -4z =34
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
cho x,y,z là các số thực dương .tìm max \(\frac{1}{4y^2+̣x+1^2+4}+\)\(\frac{1}{4z^2+y+1^2+4}+\frac{1}{4x^2+z+1^2+4}\)