Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)
\(\Rightarrow\frac{x}{3}=-4\)
\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)
\(\Rightarrow x=-12\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)
\(x+A=\frac{47}{42}\)
ta thấy :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
hay \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)
\(\frac{x}{7}\)= \(\frac{42}{45}\): \(\frac{6}{5}\)= \(\frac{7}{9}\)
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 = \(\frac{49}{9}\)
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, \(x=3\)
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, \(\frac{6+x}{33}=\frac{7}{11}\)
\(\Leftrightarrow\left(6+x\right).11=7.33\)
\(\Leftrightarrow66+11x=231\)
\(\Leftrightarrow11x=231-66\)
\(\Leftrightarrow x=\frac{165}{11}=15\)
Vậy x = 15.
b,\(\frac{12+x}{43-x}=\frac{2}{3}\)
\(\Leftrightarrow3.\left(12+x\right)=2\left(43-x\right)\)
\(\Leftrightarrow36+3x=86-2x\)
\(\Leftrightarrow3x+2x=-36+86\)
\(\Leftrightarrow5x=50\)
\(\Leftrightarrow x=10\)
Vây x = 10.
\(\frac{5}{6}\)= \(\frac{35}{42}\)
\(\Rightarrow\)x = 35 - 12
x = 23