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\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)
\(\Rightarrow\frac{x}{3}=-4\)
\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)
\(\Rightarrow x=-12\)
\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}+\frac{7}{56}+\frac{7}{72}+\frac{7}{90}\)\(\frac{7}{90}\)
=\(\frac{7}{2+6+12+20+30+42+56+72+90}\)
=\(\frac{63}{10}\)
=6.3
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) \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1\)
Vậy \(x=8\)
b.\(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
a.\(\frac{11}{14}-\frac{3}{x-1}=\frac{5}{14}\)
\(\frac{3}{x-1}=\frac{11}{14}-\frac{5}{14}=\frac{3}{7}\)
\(\Rightarrow x-1=7\)
\(x=7+1=8\)
VẬY, \(x=8\)
b. \(\frac{42}{25}:\frac{2x+1}{5}=\frac{6}{5}\)
\(\frac{2x+1}{5}=\frac{42}{25}:\frac{6}{5}=\frac{7}{5}\)
\(\Rightarrow2x+1=7\)
\(2x=7-1=6\)
\(x=6:2=3\)
VẬY, \(x=3\)
\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}=7\times\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=7\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\right)\)
\(=7\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=7\times\left(\frac{1}{1}-\frac{1}{7}\right)=7\times\left(\frac{7}{7}-\frac{1}{7}\right)=7\times\frac{6}{7}=6\)
7\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
7\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
7\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
7\(\left(1-\frac{1}{7}\right)\)
7\(.\frac{6}{7}\)
6
\(\dfrac{1}{3\cdot10}+\dfrac{1}{10\cdot17}+...+\dfrac{1}{38\cdot45}=\dfrac{6}{x}\)
=>\(\dfrac{1}{7}\left(\dfrac{7}{3\cdot10}+\dfrac{7}{10\cdot17}+...+\dfrac{7}{38\cdot45}\right)=\dfrac{6}{x}\)
=>\(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{38}-\dfrac{1}{45}=\dfrac{42}{x}\)
=>42/45=42/x
=>x=45
=>Chọn A
\(\frac{x}{7}\)= \(\frac{42}{45}\): \(\frac{6}{5}\)= \(\frac{7}{9}\)
x = 7 x \(\frac{7}{9}\)
x = \(\frac{49}{9}\)
Ta có: \(\frac{42}{45}:\frac{x}{7}=\frac{6}{5}\)
\(\Rightarrow\frac{x}{7}=\frac{42}{45}:\frac{6}{5}=\frac{7}{9}\)
=> x = 7 x 7 : 9
=> x = \(\frac{49}{9}\)
Vậy x = \(\frac{49}{9}\)