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1. a)
Vì \(\left(x-2\right).\left(y+5\right)=7\Rightarrow\)x-2 và y+5 là các ước của 7
\(Ư\left(7\right)=\left\{1;7\right\}\)
Lập bảng giá trị:
x-2 | 1 | 7 |
y+5 | 7 | 1 |
x | 3 | 9 |
y | 2 | -4 |
Chọn/Loại | Chọn | Loại |
Vậy \(x=3;y=2\)
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K NHA
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Ta có: \(\frac{x}{3}-\frac{5}{y}=\frac{1}{6}\)
\(\Leftrightarrow\frac{xy-15}{3y}=\frac{1}{6}\)
\(\Leftrightarrow6xy-90=3y\)
\(\Leftrightarrow2xy-y=30\)
\(\Leftrightarrow y\left(2x-1\right)=30\)
Đến đây xét ước là ok rồi
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\(\frac{1}{2}\times\left(x-\frac{4}{5}\right)+\frac{3}{4}x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}+\frac{3}{4}x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{3}{4}x=\frac{5}{12}+\frac{2}{5}\)
\(\Leftrightarrow\frac{5}{4}x=\frac{49}{60}\)
\(\Leftrightarrow x=\frac{49}{75}\)
Vậy \(x=\frac{49}{75}\)
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\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)
\(\Leftrightarrow x-\frac{5}{6}+x=x-\frac{2}{3}\)
\(\Leftrightarrow\left(x-x\right)+\frac{5}{6}=x-\frac{2}{3}\)
\(\Leftrightarrow0+\frac{5}{6}=x-\frac{2}{3}\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}+\frac{2}{3}\)
\(\Leftrightarrow x=\frac{5}{6}+\frac{4}{6}\)
\(\Leftrightarrow x=\frac{3}{2}\)
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