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\(a-b⋮7\Rightarrow a⋮6,b⋮7\)
\(\Rightarrow4a⋮7;3b⋮7\)
\(\Rightarrow4a+3b⋮7\) (đpcm)
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Bạn tham khảo:
Giả sử:\(\hept{\begin{cases}xyz-x=1945\left(1\right)\\xyz-y=1975\left(2\right)\\xyz-z=1995\left(3\right)\end{cases}}\)với \(x,y,z\in N\)
Tứ \(\left(1\right)\Rightarrow x\left(yz-1\right)=1945\)là số lẻ \(\Rightarrow x\)lẻ
Từ \(\left(2\right)\Rightarrow y\left(xz-1\right)=1975\)là số lẻ \(\Rightarrow y\)lẻ
Từ \(\left(3\right)\Rightarrow z\left(xy-1\right)=1995\)là số lẻ \(\Rightarrow z\)lẻ
Nên \(x,y,z\)là số lẻ
\(\Rightarrow x,y,z-x\)là số chẵn khác 1945
Vậy không tồn tại \(x,y,z\in N\)thỏa mãn \(\left(1\right),\left(2\right),\left(3\right)\).
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ta có \(10^n-1=9999...99\)(\(n-1\)chữ sô \(9\))
\(\Rightarrow10^n-1⋮9\)
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\(3\frac{1}{5}-x=1\frac{3}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{8}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{23}{10}\)
\(x=\frac{23}{10}-\frac{16}{5}\)
\(x=-\frac{9}{10}\)
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\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{48}{98}\)
\(A=\frac{8}{49}\)
A = \(\frac{1}{3}\).{ \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)}
A = \(\frac{1}{3}\).{\(\frac{1}{2}-\frac{1}{98}\)}
A = \(\frac{1}{3}.\left\{\frac{49}{98}-\frac{1}{98}\right\}\)
A=\(\frac{1}{3}.\frac{24}{49}\)
A = \(\frac{49}{98}\)
\(n^2+1⋮n-1\)
Ta có : \(n^2+1=n^2-n+n-1+2=n\left(n-1\right)+n-1+2=\left(n+1\right)\left(n-1\right)+2\)
\(\Rightarrow n^2+1⋮n-1\Leftrightarrow\left(n+1\right)\left(n-1\right)+2⋮n-1\)
\(\Leftrightarrow2⋮n-1\Leftrightarrow n-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow n\in\left\{2;0;3;-1\right\}\)