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pt \(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)+\left(\frac{x-44}{5}+3\right)=0\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
\(\Rightarrow x-29=0\Leftrightarrow x=29\)
<=> \(x\left(\frac{1}{2}+\frac{1}{y}\right)=\frac{10}{y}+\frac{3}{2}\) <=> \(x.\frac{y+2}{2y}=\frac{3y+20}{2y}\) => \(x=\frac{3y+20}{y+2}=\frac{3y+6+14}{y+2}=\frac{3\left(y+2\right)}{y+2}+\frac{14}{y+2}\) => \(x=3+\frac{14}{y+2}\)
=> để x nguyên dương thì 14 chia hết cho y+2 => y+2=(2;7;14) => y=(0,5,12) => x=(10, 5, 4), y=0 loại => Còn 2 kết quả
Đáp số: Các cặp (x, y) thỏa mãn là: (5, 7); (4, 12)
\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)
\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
=> x - 29 = 0
=> x = 29.