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\(4:x+x.2=7\)
\(\frac{1}{4}x+x.2=7\)
\(\frac{x.9}{4}=7\)
\(x=7:\frac{9}{4}=7.\frac{4}{9}=\frac{28}{9}\)
b ) \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
215 - 15 x { 25 - 15 : [ 3 x 45 - 3 x ( 50 - 2 x 3 ) ] }
= 215 - 15 x { 25 - 15 : [ 3 x 45 - 3 x 44 ] }
= 215 - 15 x { 25 - 15 : 3 }
= 215 - 15 x 20
= 215 - 300
= -85
b) \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
Ta có:
\(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}+\frac{4}{x}+...+\frac{50}{x}=255\)
\(\Rightarrow\frac{1+2+3+....+50}{x}=255\)
\(\Rightarrow\frac{\frac{50.51}{2}}{x}=255\)
\(\Rightarrow\frac{1275}{x}=255\Leftrightarrow x=\frac{1275}{255}=5\)
\(\frac{1}{x}+\frac{2}{x}+\frac{3}{x}+.........+\frac{50}{x}=255\)
\(\frac{1}{x}\left(1+2+3+....+50\right)=255\)
\(\frac{1}{x}.1275=255\)
\(\frac{1}{x}=255:1275\)
\(\frac{1}{x}=\frac{1}{5}\)
\(\Rightarrow x=5\)