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\(41-\dfrac{72}{19-108:\left[59-2\left(3^2-11\cdot2\right)^2\right]}\)

\(=41-\dfrac{72}{19-108:\left[59-2\left(18-22\right)^2\right]}\)

\(=41-\dfrac{72}{19-108:\left[59-2\cdot16\right]}\)

\(=41-\dfrac{72}{19-108:27}\)

\(=41-\dfrac{72}{19-4}=41-\dfrac{72}{15}\)

=41-4,8

=36,2

1 tháng 11 2023

\(41-72:\left\{19-108:\left[59-2.\left(3^2-11.2\right)^2\right]\right\}\)

\(=41-72:\left\{19-108:\left[59-2.\left(9-22\right)^2\right]\right\}\)

\(=41-72:\left\{19-108:\left[59-2.\left(-13\right)^2\right]\right\}\)

\(=41-72:\left\{19-108:\left[59-2.169\right]\right\}\)

\(=41-72:\left\{19-108:\left[59-338\right]\right\}\)

\(=41-72:\left\{19-108:-279\right\}\)

\(=41-72:\left\{19-\dfrac{-12}{31}\right\}\)

\(=41-72:\dfrac{601}{31}\)

\(=41-\dfrac{2232}{601}\)

\(=\dfrac{22409}{601}\)

29 tháng 12 2021

lm ơn ik có ai giúp me ko me cần gấp

29 tháng 12 2021

ai mà giúp tui thả tim hết dù đúng hay sai

6 tháng 1 2017

a,6

b,-200

c,1900

d,-900000

k mình nha 

mình cần gấp

6 tháng 1 2017

 16 . ( 38 - 2 ) - 38 . ( 16 - 1 )

=16 . 38 - 2 .16 - 38 .16 - 1.38

=16.38.(2.16-1.38)

=608.(32-38)

=608.(-6)

=-3648

Ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)

\(=8-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

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

\(=8-1+\dfrac{1}{10}\)

\(=\dfrac{71}{10}\)

25 tháng 8 2023

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

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

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

\(=8-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\) ( có 8 số hạng 1)

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

\(=8-\left(1-\dfrac{1}{10}\right)\)

\(=8-\dfrac{9}{10}\)

\(=\dfrac{80}{10}-\dfrac{9}{10}=\dfrac{71}{10}\)

25 tháng 8 2023

A=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90

=1−1/2+1−1/6+1−1/12+1−1/20+1−1/30+1−1/42+1−1/56+1−1/72+1−1/90

=9−(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)

=9−(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)

=9-(1-1/2+1/2-1/3+.....+1/9-1/10)

=9−(1−1/10)

=9−1+1/10=8+1/10=81/10

12 tháng 6 2016

\(9-A=1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+1-\frac{19}{20}+...+1-\frac{89}{90}\)

\(\Leftrightarrow9-A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Leftrightarrow9-A=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{10-9}{9\cdot10}\)

\(\Leftrightarrow9-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{10}\)

14 tháng 5 2016

1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/20)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)=(1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72)=8-(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)=8-(1-1/2+1/2-1/3+...+1/8-1/9)=8-(1-1/9)=8-8/9=72/9-8/9=64/9

1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72

=10-(1/2+1/6+..+1/110)

=10-(1/1x2+1/2x3+...+1/10x11)

=10-(1-1/2+1/2-1/3+...+1/10-1/11)

=10-(1-1/11)

=10-10/11

=100/11