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21 tháng 7 2023

\(...\dfrac{4}{6}+\dfrac{88}{12}=\dfrac{2}{3}+\dfrac{22}{3}=\dfrac{24}{3}=8\)

21 tháng 7 2023

8

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}\)

\(=10-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)

\(=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

=9+1/10

=9,1

24 tháng 8 2023

a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)

b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)

c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)

10 tháng 4 2017

\(Q=\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)

\(Q=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)+\left(1-\dfrac{1}{72}\right)+\left(1-\dfrac{1}{90}\right)\)

\(Q=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(Q=9-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(Q=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(Q=9-\left(1-\dfrac{1}{10}\right)\)

\(Q=9-\dfrac{9}{10}\)

\(Q=\dfrac{81}{10}\)

Chúc bạn học tốt :))

29 tháng 8 2017

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+1-\dfrac{1}{30}+1-\dfrac{1}{42}+1-\dfrac{1}{56}+1-\dfrac{1}{72}+1-\dfrac{1}{90}\)

\(=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=9-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=9-\left(1-\dfrac{1}{10}\right)\)

\(=9-\dfrac{9}{10}\)

\(=\dfrac{81}{10}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

31 tháng 12 2022

a: \(\Leftrightarrow4x+4+9\left(2x+1\right)=2\left(5x+3\right)+12x+7\)

=>4x+4+18x+9=10x+6+12x+7

=>22x+13=22x+13(luôn đúng)

b: \(\Leftrightarrow\left(\dfrac{x+1}{94}+1\right)+\left(\dfrac{x+2}{93}+1\right)+\left(\dfrac{x+3}{92}+1\right)=\left(\dfrac{x+4}{91}+1\right)+\left(\dfrac{x+5}{90}+1\right)+\left(\dfrac{x+6}{89}+1\right)\)

=>x+95=0

=>x=-95

9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)

10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)

\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0