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3 tháng 8 2017

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+....+\dfrac{3}{32\cdot35}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{32}-\dfrac{1}{35}\)

\(=\dfrac{1}{2}-\dfrac{1}{35}\)

\(=\dfrac{33}{70}\)

3 tháng 8 2017

thank youvui

9 tháng 5 2023

A =\(\dfrac{4^2}{3\times5}\) \(\times\)\(\dfrac{5^2}{4\times6}\) \(\times\) \(\dfrac{6^2}{5\times7}\) \(\times\) \(\dfrac{7^2}{6\times8}\)

A = \(\dfrac{4\times4\times5^2\times6^2\times7\times7}{3\times4\times5^2\times6^2\times7\times8}\)

A =  \(\dfrac{4}{3}\) \(\times\) \(\dfrac{7}{8}\)

A = \(\dfrac{7}{6}\) 

21 tháng 2 2023

a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)

b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)

c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)

21 tháng 2 2023

giúp mik với ạ, mình sẽ tick ạ! Thanks.

2 tháng 4 2018

\(B=\dfrac{40404}{70707}+\dfrac{244\times395-151}{244+395\times243}+\dfrac{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}{1\times5\times7+2\times10\times14+4\times20\times28+7\times35\times49}\\ =\dfrac{4}{7}+\dfrac{243\times395+395-151}{244+395\times243}+\dfrac{1\times3\times5\left(1+2+4+7\right)}{1\times5\times7\left(1+2+4+7\right)}\\ =\dfrac{4}{7}+\dfrac{243\times395+244}{244+395\times243}+\dfrac{3}{7}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+1\\ =1+1=2\)

24 tháng 1 2022

\(\dfrac{2}{35}\)

9 tháng 3 2017

a,

\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)

b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)

c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)

7 tháng 9 2021

\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{32}{40}+\dfrac{48}{56}+\dfrac{14}{21}\\ =\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{6}{7}+\dfrac{2}{3}\\ =\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{7}+\dfrac{6}{7}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\\ =1+1+1=3\)

AH
Akai Haruma
Giáo viên
7 tháng 9 2021

Lời giải:

$\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{32}{40}+\frac{48}{56}+\frac{14}{21}$

$=\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{4}{5}+\frac{6}{7}+\frac{2}{3}$

$=(\frac{1}{3}+\frac{2}{3})+(\frac{1}{7}+\frac{6}{7})+(\frac{1}{5}+\frac{4}{5})$

$=\frac{3}{3}+\frac{7}{7}+\frac{5}{5}=1+1+1=3$

23 tháng 9 2018

Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)

=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)

=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)

=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)

=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)

8 tháng 1 2023

a)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{7}{2}\)

\(=\dfrac{13}{5}+\dfrac{49}{10}\\ =\dfrac{26}{10}+\dfrac{49}{10}\\ =\dfrac{15}{2}\)

b)

\(=\dfrac{52}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{52}{4}-\dfrac{22}{7}\\ =\dfrac{69}{7}\)

TC
Thầy Cao Đô
Giáo viên VIP
8 tháng 1 2023

a) $2\dfrac35 + 1\dfrac25 . 3\dfrac12$

$= \dfrac{13}5 + \dfrac75.\dfrac72$

$= \dfrac{26}{10} + \dfrac{49}{10}$

$=\dfrac{15}2$.

b) $4\dfrac34 - 3\dfrac23 : 1\dfrac16$

$= \dfrac{19}4 - \dfrac{11}3 : \dfrac76$

$= \dfrac{19}4 - \dfrac{11}3 . \dfrac67$

$= \dfrac{19}4 - \dfrac{22}7$

$= \dfrac{45}{28}$.