Tính :
\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right).\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
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TT
31 tháng 8 2017
a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)
= \(\dfrac{3}{11}\)
b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
= \(\dfrac{3}{5}\)
c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)
= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)
= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)
= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)
PN
1 tháng 1 2018
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{3862}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
\(=1:5\)
\(=\dfrac{1}{5}\)
\(=0,2\)
AN
1 tháng 1 2018
Theo đề ta có:
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\left(\dfrac{4}{368}-\dfrac{3}{368}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{2}.\dfrac{1}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{34}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{14}{3862}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=>\(1:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[1+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{2}{2}+\dfrac{9}{2}\right]\)
=> \(1:\dfrac{11}{2}\)
=> \(1.\dfrac{2}{11}\)
=> \(\dfrac{2}{11}\)
28 tháng 9 2018
Đây là tính hợp lí ... mà câu a là 27,5 chứ không phải 2,75...
\(A=\dfrac{7,5-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{27,5-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{15}{2}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{55}{2}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{3}}\\ =\dfrac{3\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
8 tháng 9 2022
b: \(=26:\left[\dfrac{3:0.1}{2.5\cdot2}+\dfrac{0.25\cdot4}{2}\right]+\dfrac{2}{3}\cdot\dfrac{21}{4}\)
\(=26:\left[\dfrac{30}{5}+1\right]+\dfrac{42}{12}\)
\(=\dfrac{26}{7}+\dfrac{42}{12}=\dfrac{101}{14}\)
c: \(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{\left(\dfrac{1}{34}+\dfrac{33}{34}\right)}{\dfrac{1}{2}+\dfrac{9}{2}}=1:5=\dfrac{1}{5}\)
31 tháng 12 2023
a: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{1}{2}\left(2+3+4+...+17\right)\)
\(=\dfrac{1}{2}\cdot152=76\)
b: Sửa đề: \(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right)\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{193}\cdot\dfrac{193}{17}-\dfrac{3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left[\dfrac{7}{1931}\cdot\dfrac{1931}{25}+\dfrac{11}{3862}\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)
\(=\left(\dfrac{2}{17}+\dfrac{30}{34}\right):\dfrac{14+11+225}{50}\)
\(=1\cdot\dfrac{50}{250}=1\cdot\dfrac{1}{5}=\dfrac{1}{5}\)
c: Sửa đề: \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
d: \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)
\(=\dfrac{1}{3}+1:\dfrac{3}{2}=1\)
EV
4 tháng 7 2016
\(M=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right] \)
\(=\left(\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right):\left(\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right)\)
\(=\frac{4-3+33}{34}:\frac{14+11+225}{50}=1:5=0.2\)
25 tháng 8 2018
Ta có : [ ( 2/193 - 3/386 ) * 193/17 + 32/34 ] : [ ( 7/2001 + 11/4002 ) * 2001/25 + 9/2 ] .
=> [ 2/193 * 193/17 - 3/386 * 193/17 + 32/34 ] : [ 7/2001 * 2001/25 + 11/4002 * 2001/25 + 9/2 ] .
=> [ 2/17 - 3/34 + 32/34 ] : [ 7/25 + 11/50 + 9/2 ] .
=> [ 4/34 - 3/34 + 32/34 ] : [ 14/50 + 11/50 + 225/50 ] .
=> 33/34 : 5 .
=> 33/34 * 1/5 .
=> 33/170 .
LL
25 tháng 8 2018
Tính bằng cách thuận tiện nhất:
\(=\left[\left(\frac{2}{193}\cdot\frac{193}{17}\right)-\left(\frac{3}{386}\cdot\frac{193}{17}\right)+\frac{32}{34}\right]:\left[\left(\frac{7}{2001}\cdot\frac{2001}{25}\right)+\left(\frac{11}{4002}\cdot\frac{2001}{25}\right)+\frac{9}{2}\right]\)
\(=\left[\left(\frac{2}{17}-\frac{3}{17}\right)+\frac{32}{34}\right]:\left[\left(\frac{7}{25}+\frac{11}{50}\right)+\frac{9}{2}\right]\)
\(=\left(-\frac{1}{17}+\frac{32}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=\frac{15}{17}+5\)
\(=\frac{100}{17}\)
~ học tốt ~
3 tháng 2 2022
\(M=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{14+11}{4002}-\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{1}{17}\cdot\dfrac{193}{386}+\dfrac{33}{34}\right):\left[\dfrac{25}{4002}-\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=1:\dfrac{625-2001\cdot4002+9\cdot50525}{100050}\)
\(=-\dfrac{100050}{7552652}\)
9 tháng 3 2022
d: \(=\dfrac{-7}{9}\left(\dfrac{3}{11}+\dfrac{8}{11}\right)+1+\dfrac{7}{9}=1\)
e: \(=\dfrac{1}{5}\left(\dfrac{10}{19}+\dfrac{9}{19}\right)-\dfrac{2}{35}=\dfrac{1}{5}-\dfrac{2}{35}=\dfrac{5}{35}=\dfrac{1}{7}\)
f: \(=\left(-25\cdot4\right)\cdot\left(-8\cdot125\right)\cdot\left(-17\right)=-1700000\)
ta có
\(M=[(\dfrac{2}{193}-\dfrac{3}{386}).\dfrac{193}{17}+\dfrac{33}{34}]:[(\dfrac{7}{2001}+\dfrac{11}{4002}).\dfrac{2001}{25}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}]:[\dfrac{25}{4002}.\dfrac{2001}{25}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=[\dfrac{1}{34}+\dfrac{33}{34}]:[\dfrac{1}{2}+\dfrac{9}{2}]\)
\(\Rightarrow\)\(M=1:5\)
\(\Rightarrow M=\dfrac{1}{5}\)