Cho x,y,z khác 0 và x-2y+3z tính N=(1+3 z/x)(2-x/y)(3-2y/z)
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2,46 : 4 + 2,46 x 0,75
= 2,46 x 0,25 + 2,46 x 0,75
= 2,46 x ( 0,25 + 0,75 )
= 2,46 x 1
= 2,46
\(2,46:4+2,46\times0,75\)
\(=2,46:\frac{4}{1}+2,46\times0,75\)
\(=2,46\times\frac{1}{4}+2,46\times0,75\)
\(=2,46\times\left[\frac{1}{4}+0,75\right]\)
\(=2,46\times\left[0,25+0,75\right]=2,46\)
\(\frac{50}{51}\cdot\frac{51}{15}:\frac{34}{15}\)
\(=\frac{50}{51}\cdot\frac{51}{15}\cdot\frac{15}{34}=\frac{50\cdot51\cdot15}{51\cdot15\cdot34}=\frac{50}{34}=\frac{25}{17}\)
\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{3}{4}\)
\(\left[\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right]:\frac{5}{6}+\frac{1}{2}\)
\(=\left[\frac{3\cdot3}{24}+\frac{-3\cdot6}{24}+\frac{7\cdot2}{24}\right]:\frac{5}{6}+\frac{1}{2}\)
\(=\left[\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right]:\frac{5}{6}+\frac{1}{2}\)
\(=\left[\frac{9+(-18)+14}{24}\right]:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}\cdot\frac{6}{5}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{1}{4}+\frac{1\cdot2}{4}=\frac{1}{4}+\frac{2}{4}=\frac{3}{4}\)
Theo đề bài,đặt \(x+y=k\inℤ\) (1)
\(\frac{1}{x}+\frac{1}{y}=\left(x+y\right).\frac{1}{xy}=k.\frac{1}{xy}\)
Do k nguyên (theo (1)) nên để \(\frac{1}{x}+\frac{1}{y}\) nguyên thì \(\frac{1}{xy}\) nguyên
Nên \(xy\inƯ\left(1\right)=\left\{1;-1\right\}\)
Suy ra \(\left(x;y\right)=\left(1;1\right),\left(-1;-1\right),\left(1;-1\right),\left(-1;1\right)\)
Đúng không ta?