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![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2019}{y}=\frac{x+y-2020}{z}=\frac{y+z+1+x+z+2019+x+y-2020}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow2=\frac{1}{x+y+z}\)\(\Rightarrow x+y+z=\frac{1}{2}\)
Ta có:
+) \(\frac{y+z+1}{x}=2\)\(\Rightarrow y+z+1=2x\)\(\Rightarrow x+y+z+1=3x\)\(\Rightarrow\frac{1}{2}+1=3x\)\(\Rightarrow3x=\frac{3}{2}\)\(\Rightarrow x=\frac{1}{2}\)
+) \(\frac{x+z+2019}{y}=2\)\(\Rightarrow x+z+2019=2y\)\(\Rightarrow x+y+z+2019=3y\)\(\Rightarrow\frac{1}{2}+2019=3y\)\(\Rightarrow3y=\frac{4039}{2}\)\(\Rightarrow y=\frac{4039}{6}\)
+) \(\frac{x+y-2020}{z}=2\)\(\Rightarrow x+y-2020=2z\)\(\Rightarrow x+y+z-2020=3z\)\(\Rightarrow\frac{1}{2}-2020=3z\)\(\Rightarrow3z=\frac{-4039}{2}\)\(\Rightarrow z=\frac{-4039}{6}\)
Lại có: \(A=2016x+y^{2017}+z^{2017}=2016.\frac{1}{2}+\left(\frac{4039}{6}\right)^{2017}+\left(\frac{-4039}{6}\right)^{2017}=4032+\left(\frac{4039}{6}\right)^{2017}-\left(\frac{4039}{6}\right)^{2017}=4032\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{-\left(-x\right)}{5}-\frac{2}{10}=\frac{1}{-5}-\frac{7}{50}\)
\(\frac{x}{5}-\frac{2}{10}=\frac{-1}{5}-\frac{7}{50}\)
\(\frac{x}{5}-\frac{2}{10}=-\frac{1}{5}+-\frac{7}{50}\)
\(\frac{x}{5}-\frac{2}{10}=-\frac{17}{50}\)
\(\frac{x}{5}=-\frac{17}{50}+\frac{2}{10}\)
\(\frac{x}{5}=-\frac{7}{50}\)
\(\Rightarrow x=-\frac{7}{50}.5\)
Vậy \(x=-\frac{7}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời hộ mik đi các bn, trả lời xong mik kik cho
Ta có: \(\frac{1}{x}-\frac{y}{8}=\frac{1}{16}\)
=> \(\frac{1}{x}=\frac{1}{16}+\frac{y}{8}\)
=> \(\frac{1}{x}=\frac{1+2y}{16}\)
=> 1.16 = x(1 + 2y)
=> x(1 + 2y) = 16 = 1 . 16 = 2 . 8 = 4.4
Vì 1 + 2y là số lẽ nên 1 + 2y \(\in\){1; -1} => x \(\in\){16; -16}
Lập bảng :
Vậy ...
:
1x =116
=> =>
X = 1.16:1 =16
Y=1.8:16= 0.5
y8 =116
Vậy X = 16 ; Y=0.5
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