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\(5x=4y=2z\)
\(\Leftrightarrow\frac{5x}{20}=\frac{4y}{20}=\frac{2z}{20}\)
\(\Leftrightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{10}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}=\frac{y}{5}=\frac{z}{10}=\frac{x-y+z}{4-5+10}=\frac{-18}{8}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{4}=-2\\\frac{y}{5}=-2\\\frac{z}{10}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=-10\\z=-20\end{matrix}\right.\)
Lại có :
\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}\)
\(=\left(\frac{2}{-8}+\frac{5}{-10}+\frac{5}{-20}\right)^{2016}\)
\(=1\)
Vậy....
Đặt 5x=4y=2z=k suy ra \(x=\frac{k}{5};y=\frac{k}{4};z=\frac{k}{2}\)
Ta có :
x-y+z=-18
\(\frac{k}{5}-\frac{k}{4}+\frac{k}{2}=-18\)
\(k.\left(\frac{1}{5}-\frac{1}{4}+\frac{1}{2}\right)=-18\)
\(k.\frac{9}{20}=-18\)
k = -40 suy ra x = -8 ; y = -10 ; z = -20
Ta có:
\(A=\left(\frac{2}{x}+\frac{5}{y}+\frac{5}{z}\right)^{2016}=\left(\frac{2}{-8}+\frac{5}{-40}+\frac{5}{-20}\right)^{2016}=\left(\frac{5}{-8}\right)^{2016}=0\)
Bài 1 :
\(3x+5=2\left(x-\frac{1}{4}\right)\)
\(\Leftrightarrow3x+5=2x-\frac{1}{2}\)
\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)
\(\Leftrightarrow\frac{11}{2}=-x\)
\(\Leftrightarrow\frac{-11}{2}=x\)
Vậy \(x=\frac{-11}{2}\)
Bài 2:
a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)
\(\Leftrightarrow x+\frac{19}{5}=0\)
\(\Leftrightarrow x=\frac{-19}{5}\)
\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)
\(\Leftrightarrow y+\frac{2018}{2019}=0\)
\(\Leftrightarrow y=\frac{-2018}{2019}\)
\(\Rightarrow+,\left|z-3\right|=0\)
\(\Leftrightarrow z-3=0\)
\(\Leftrightarrow z=3\)
Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)
b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)
Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)
\(\Rightarrow x\inℚ\)
\(\Rightarrow+,\left|2y+4\right|\ge0\)
\(\Rightarrow y\inℚ\)
\(\Rightarrow+,\left|z-5\right|\ge0\)
\(\Rightarrow z\inℚ\)
Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.
1.4m+7n=0
=>4m=-7n
=>mx2-4m=0
=>m(x2-4)=0
=>m=0 hoặc x=2 hoặc x=-2
Ta có :\(\frac{x}{4y+z}=\frac{y}{4z+x}=\frac{z}{4x+y}=\frac{x+y+z}{4y+z+4z+x+4x+y}=\frac{x+y+z}{5\left(x+y+z\right)}=\frac{1}{5}\)
=> \(\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{y}{4z+x}=\frac{1}{5}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{4z+x}{y}=5\end{cases}}\)
Khi đó A = 2019 - 1/5 + 5 = 2023,8
\(\frac{x}{4y+z}=\frac{y}{4z+x}=\frac{z}{4x+y}=\frac{x+y+z}{4y+z+4z+x+4x+y}=\frac{x+y+z}{5\left(x+y+z\right)}=\frac{1}{5}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{y}{4z+x}=\frac{1}{5}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{4z+x}{y}=5\end{cases}}}\)
Khi đó \(A=2019-\frac{1}{5}+5=2013,8\)
\(\frac{3x-2y}{2015}=\frac{2x-4x}{2016}=\frac{4y-3z}{2017}\)
\(\Rightarrow\frac{12x-8y}{8060}=\frac{6z-12x}{6048}=\frac{8y-6z}{4034}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{8060+6048+4034}=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-2y=0\\2z-4x=0\\4y-3z=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\2z=4x\\4y=3z\end{cases}}}\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\\\frac{y}{3}=\frac{z}{4}\end{cases}}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\left(k\ne0\right)\)
\(\Rightarrow x=2k;y=3k;z=4k\)
Thay vào P ta có
\(P=\frac{4k^2-2.2k.3k-16k^2}{4k^2+9k^2+16k^2}=\frac{k^2\left(4-12-16\right)}{k^2\left(4+9+16\right)}=-\frac{24}{29}\)