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17 tháng 3 2018
TH1 : \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)
Th2 : \(x+y+z\ne0\)
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)
\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)
\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)
\(\Leftrightarrow x=y=z\)
\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)
Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)
LD
14 tháng 1 2019
Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
AT
2
G
20 tháng 9 2017
Giải:
Theo đề ra, ta có:
\(\dfrac{x-y}{x+2y}=\dfrac{3}{4}\)
\(\Leftrightarrow3.\left(x+2y\right)=4.\left(x-y\right)\)
\(\Leftrightarrow3x+6y=4x-4y\)
\(\Leftrightarrow6y+4y=4x-3x\)
\(\Leftrightarrow10y=x\)
\(\Leftrightarrow\dfrac{x}{y}=10\)
Vậy ...
Chúc bạn học tốt!
20 tháng 9 2017
Dễ thấy \(y\ne0\)
Ta có: \(\dfrac{x-y}{x+2y}=\dfrac{3}{4}\Leftrightarrow\dfrac{\dfrac{x}{y}-\dfrac{y}{y}}{\dfrac{x}{y}+\dfrac{2y}{y}}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\dfrac{x}{y}-1}{\dfrac{x}{y}+2}=\dfrac{3}{4}\)
Đặt \(\dfrac{x}{y}=k\) với k là số hữu tỉ
Khi đó có \(\dfrac{k-1}{k+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{k+2}{4}=\dfrac{k-1}{3}=\dfrac{k+2-\left(k-1\right)}{4-3}=3\)
\(\Rightarrow k=10\)
Vậy \(\dfrac{x}{y}=10\)
16 tháng 3 2017
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\dfrac{x}{2x+y+z}\)=\(\dfrac{y}{2y+x+z}\)=\(\dfrac{z}{2z+x+y}\)=\(\dfrac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}\)=\(\dfrac{x+y+z}{3x+3y+3z}\)=\(\dfrac{x+y+z}{3.\left(x+y+z\right)}\)=\(\dfrac{1}{3}\)=\(\dfrac{3}{9}\)<\(\dfrac{3}{4}\)(đpcm)
29 tháng 5 2022
a: \(\dfrac{2x-y}{3x+2y}=\dfrac{5}{2}\)
\(\Leftrightarrow15x+10y=4x-2y\)
=>11x=-12y
=>\(\dfrac{x}{-12}=\dfrac{y}{11}\)
Đặt \(\dfrac{x}{-12}=\dfrac{y}{11}=k\)
=>x=-12k; y=11k
\(P=\dfrac{5x+4y}{25x-y}=\dfrac{5\cdot\left(-12k\right)+4\cdot11k}{25\cdot\left(-12k\right)-11k}=\dfrac{16}{311}\)
b: \(\dfrac{x-5y}{x-3y}=\dfrac{4}{3}\)
=>4x-12y=3x-15y
=>x=-3y
\(\Leftrightarrow\dfrac{x}{-3}=\dfrac{y}{1}=k\)
=>x=-3k; y=k
\(P=\dfrac{x^3+2y^3}{x^3-y^3}=\dfrac{-27k^3+2k^3}{-27k^3-k^3}=\dfrac{-25}{-28}=\dfrac{25}{28}\)
LK
Cho \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(A=\dfrac{x+y-z}{x-y+z};B=\dfrac{2x+3y+z}{x-2y-3z}\)
2
7 tháng 8 2017
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow x=2k;y=3k;z=4k\)
sau đó bạn tự thay vào A và B r tính nhá
7 tháng 8 2017
Đặt:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{2k+3k-4k}{2k-3k+4k}=\dfrac{k}{3k}=\dfrac{1}{3}\)
\(\Rightarrow B=\dfrac{2.2k+3.3k+4k}{2k-2.3k-3.4k}=\dfrac{4k+9k+4k}{2k-6k-12k}=\dfrac{17k}{-16k}=\dfrac{17}{-16}\)
18 tháng 4 2017
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
5 tháng 11 2017
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
5 tháng 11 2017
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
\(\dfrac{2x+y}{x-2y}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(2x+y\right)=3\left(x-2y\right)\)
\(\Leftrightarrow8x+4y=3x-6y\)
\(\Leftrightarrow4y+6y=3x-8x\)
\(\Leftrightarrow10y=-5x\)
\(\Leftrightarrow\dfrac{x}{y}=-2\)
Vậy..