TÌm x ; y ; z sao cho :
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{5}\) ( x , y , z ∈ N* )
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
Cho \(x,y,z\) dương sao cho \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=6\). Tìm giá trị lớn nhất của \(P=\dfrac{1}{3x+3y+2z}+\dfrac{1}{3y+3z+2x}+\dfrac{1}{3z+3x+2y}\)
Cho x,y,z là các số dương thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). Tìm Max \(F=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)
Áp dụng BĐT BSC:
\(F=\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)
\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)
Tìm x, y, z, t ∈ Z biết:
a, \(\dfrac{5}{x}=\dfrac{-10}{12}\) b, \(\dfrac{4}{-6}=\dfrac{x+3}{9}\) c, \(\dfrac{x-1}{25}=\dfrac{4}{x-1}\) d, \(\dfrac{x+1}{y}=\dfrac{-3}{5}\)
e, \(\dfrac{-12}{6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{Z}{-17}=\dfrac{-t}{-9}\)
h, \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{Z^3}{-2}\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
cho x , y, z ≠0 thỏa mãn \(\dfrac{x+y-z}{z}\)=\(\dfrac{y+z-x}{x}\)=\(\dfrac{z+x-y}{y}\). tính P=(1+\(\dfrac{x}{y}\)).(1 +\(\dfrac{y}{z}\)).(1+\(\dfrac{z}{x}\))
Lời giải:
Nếu $x+y+z=0$ thì:
$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$
$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$
$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$
(thỏa mãn đkđb)
Khi đó:
$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$
$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$
Nếu $x+y+z\neq 0$
Áp dụng TCDTSBN:
$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$
$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:
$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$
cho x,y,z là các số nguyên dương với \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)
Tìm max : \(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+xz}+\dfrac{z}{z^2+xy}\)
\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = z = 1.
Tìm x,y,z:
\(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)
Lời giải:
Áp dụng TCDTSBN:
$\frac{1}{x+y+z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2(x+y+z)}{x+y+z}=2$
\(\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ x+y+z+1=3x\\ x+y+z+2=3y\\ x+y+z-3=3z\end{matrix}\right.\)
\(\left\{\begin{matrix} \frac{1}{2}+1=3x\\ \frac{1}{2}+2=3y\\ \frac{1}{2}-3=3z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{5}{6}\\ z=\frac{-5}{6}\end{matrix}\right.\)
Cho 3 số thực x,y,z thỏa mãn \(\dfrac{1}{x^{2}} + \dfrac{1}{y^{2}} + \dfrac{1}{z^{2}}\)= 3
Tìm GTNN của biểu thức P = \(\dfrac{y^{2}z^{2}}{x(y^{2}+z^{2})} + \dfrac{z^{2}x^{2}}{y(z^{2}+x^{2})} + \dfrac{x^{2}y^{2}}{z(x^2+y^2)}\)
Lời giải:
Bạn cần bổ sung điều kiện $x,y,z>0$
\(P=\frac{1}{x.\frac{y^2+z^2}{y^2z^2}}+\frac{1}{y.\frac{z^2+x^2}{z^2x^2}}+\frac{1}{z.\frac{x^2+y^2}{x^2y^2}}=\frac{1}{x(\frac{1}{y^2}+\frac{1}{z^2})}+\frac{1}{y(\frac{1}{z^2}+\frac{1}{x^2})}+\frac{1}{z(\frac{1}{x^2}+\frac{1}{y^2})}\)
\(=\frac{1}{x(3-\frac{1}{x^2})}+\frac{1}{y(3-\frac{1}{y^2})}+\frac{1}{z(3-\frac{1}{z^2})}=\frac{x}{3x^2-1}+\frac{y}{3y^2-1}+\frac{z}{3z^2-1}\)
Vì $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3\Rightarrow x^2, y^2, z^2>\frac{1}{3}$
Xét hiệu:
\(\frac{x}{3x^2-1}-\frac{1}{2x^2}=\frac{(x-1)^2(2x+1)}{2x^2(3x^2-1)}\geq 0\) với mọi $x>0$ và $x^2>\frac{1}{3}$
$\Rightarrow \frac{x}{3x^2-1}\geq \frac{1}{2x^2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế ta có:
$P\geq \frac{1}{2}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{3}{2}$
Vậy $P_{\min}=\frac{3}{2}$ khi $x=y=z=1$
Tìm x;y;z biết
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\dfrac{1}{x+y+z}=2\Rightarrow2x+2y+2z=1\Rightarrow x+y+z=0,5\Rightarrow\left\{{}\begin{matrix}x+y=0,5-z\\y+z=0,5-x\\x+z=0,5-y\end{matrix}\right.\\ \dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow0,5-x+1=2x\Rightarrow x=0,5\\ \dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow0,5-y+2=2y\Rightarrow y=\dfrac{5}{6}\\ \dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\Rightarrow0,5-z-3=2z\Rightarrow z=-\dfrac{5}{6}\)