K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

21 tháng 10 2021

\(18\dfrac{2}{7}.\dfrac{3}{4}-6\dfrac{2}{7}.\dfrac{3}{4}=\dfrac{128}{7}.\dfrac{3}{4}-\dfrac{44}{7}.\dfrac{3}{4}=\left(\dfrac{128}{7}-\dfrac{44}{7}\right).\dfrac{3}{4}=12.\dfrac{3}{4}=9\)

21 tháng 10 2021

\(18\dfrac{2}{7}\cdot\dfrac{3}{4}-6\dfrac{2}{7}\cdot\dfrac{3}{4}\)

\(=\dfrac{3}{4}\left(18+\dfrac{2}{7}-6-\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{3}{4}=9\)

21 tháng 10 2021

\(18\dfrac{2}{7}\cdot\dfrac{3}{4}-6\dfrac{2}{7}\cdot\dfrac{3}{4}\)

\(=\dfrac{3}{4}\left(18+\dfrac{2}{7}-6-\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{3}{4}=9\)

20 tháng 12 2021

Cho mik cái này đề bài vs

20 tháng 12 2021

e: \(=\dfrac{1}{6}-\dfrac{4}{9}+\dfrac{5}{18}=\dfrac{3-8+5}{18}=0\)

20 tháng 12 2022

g)\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right).\dfrac{7}{3}\)
   \(=\left(-\dfrac{3}{4}+-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right).\dfrac{7}{3}\)
\(=\left(-1+1\right).\dfrac{7}{3}=0.\dfrac{7}{3}=0\)

20 tháng 12 2022

f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)

\(=\dfrac{3^3.5^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)

\(=\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)

\(=\dfrac{5^3}{6^3}\)

\(=\dfrac{125}{216}\)

22 tháng 12 2021

\(a,=\dfrac{9}{12}-\dfrac{7}{12}-\dfrac{8}{12}=-\dfrac{1}{2}\\ b,=\dfrac{95}{6}+\dfrac{17}{18}-\dfrac{5}{12}=\dfrac{589}{36}\)

a: \(=6+\dfrac{4}{5}-1-\dfrac{2}{3}-3-\dfrac{4}{5}\)

\(=2-\dfrac{2}{3}=\dfrac{4}{3}\)

b: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

c: =6+7/7-1-3/4-2-5/7

=3+2/7-3/4

=84/28+8/28-21/28

=84/28-13/28

=71/28

31 tháng 12 2023

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

11 tháng 11 2018

a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)

= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)

= 0 + \(\dfrac{11}{125}\)

= \(\dfrac{11}{125}\)

b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +

\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4

= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4

= -1

c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)

= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)

= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)

= \(\dfrac{-7}{75}\)

d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)

= 1 + (-1)

= 0

\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)