\(\frac{1}{x}+\frac{y}{3}=\frac{1}{6}\)
=> \(\frac{1}{x}=\frac{1}{6}-\frac{y}{3}\)
=> \(\frac{1}{x}=\frac{1-2y}{6}\)
=> \(x\left(1-2y\right)=6\)
=> \(x;1-2y\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
Vì \(y\in N\Rightarrow1-2y\in\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{2;6\right\}\)
Lập bảng :
| 1 - 2y | 1 | 3 |
| x | 6 | 2 |
| y | 0 | -1 (loại) |
Vậy ...
\(\frac{3}{4}-2.\left|2x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow2.\left|2x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow2.\left|2x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\left|2x-\frac{2}{3}\right|=\frac{1}{4}:2\)
\(\Rightarrow\left|2x-\frac{2}{3}\right|=\frac{1}{8}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{2}{3}=\frac{1}{8}\\2x-\frac{2}{3}=\frac{-1}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{8}+\frac{2}{3}\\2x=\frac{-1}{8}+\frac{2}{3}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{19}{24}\\2x=\frac{13}{24}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{19}{24}:2\\x=\frac{13}{24}:2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{48}\\x=\frac{13}{48}\end{cases}}\)
Vậy ...................................
~ Hok tốt ~
1. Ta có:
\(\frac{1}{x}+\frac{y}{3}=\frac{1}{6}\Rightarrow\frac{1}{x}=\frac{1}{6}-\frac{y}{3}\)
\(\Rightarrow\frac{1}{x}=\frac{1-2y}{6}\)
\(\Rightarrow x.\left(1-2y\right)=6\)
\(\Rightarrow x;1-2y\inƯ\left(6\right)\)
Mà 1 - 2y là số lẻ => 1 - 2y thuộc {-1;-3;1;3}
Ta có bảng:
| 1 - 2y | -1 | -3 | 1 | 3 |
| x | -6 | -2 | 6 | 2 |
| y | 1 | 3/2 | 0 | -1 |
2.
a) \(\frac{3}{4}-2\times\left|2x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow2\times\left|2x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow\left|2x-\frac{2}{3}\right|=\frac{1}{4}:2=\frac{1}{8}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{2}{3}=\frac{1}{8}\\2x-\frac{2}{3}=\frac{-1}{8}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{8}+\frac{2}{3}\\2x=\frac{-1}{8}+\frac{2}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{19}{24}\\2x=\frac{13}{24}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{24}:2\\x=\frac{13}{24}:2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{19}{48}\\x=\frac{13}{48}\end{cases}}\)
b) \(\left(2x+\frac{3}{5}\right)^2+\frac{9}{25}=1\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}=\left(\frac{4}{5}\right)^2=\left(\frac{-4}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{4}{5}\\2x+\frac{3}{5}=\frac{-4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{5}\\2x=\frac{-7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}\\x=\frac{-7}{10}\end{cases}}\)
