Bài 1 :

\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< ..................< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< .....................< \dfrac{1}{48}-\left(\dfrac{-5}{48}\right)\)

\(\Leftrightarrow\dfrac{5}{12}< .............< \dfrac{1}{8}\)

\(\Leftrightarrow0,41\left(6\right)< ...........< 0,125\)

\(\Leftrightarrow\) ko tìm dc số nguyên thích hợp vào chỗ chấm

1) $(-81)\cdot27-3^4\cdot(-77)$

$=(-81)\cdot27-81\cdot(-77)$

$=-81\cdot[27+(-77)]$

$=-81\cdot(27-77)$

$=-81\cdot(-50)$

$=81\cdot50$

$=4050$

2) $(-16)\cdot32-(-4)^2\cdot18$

$=(-16)\cdot32-16\cdot18$

$=-16\cdot(32+18)$

$=-16\cdot50$

$=-800$

3) $(-2)^4\cdot289-16\cdot189$

$=16\cdot289+16\cdot(-189)$

$=16\cdot[289+(-189)]$

$=16\cdot(289-189)$

$=16\cdot100$

$=1600$

4) $41\cdot(-18)+41\cdot(-81)-41$

$=1\cdot(-18)+41\cdot(-81)+41\cdot(-1)$

$=41\cdot[(-18)+(-81)+(-1)]$

$=41\cdot(-18-81-1)$

$=41\cdot(-99-1)$

$=41\cdot(-100)$

$=-4100$

5) $17\cdot(-37)-23\cdot37-46\cdot(-37)$

$=17\cdot(-37)+23\cdot(-37)-46\cdot(-37)$

$=-37\cdot(17+23-46)$

$=-37\cdot(40-46)$

$=-37\cdot(-6)$

$=37\cdot6$

$=222$

6) $-48+48\cdot(-78)+48\cdot(-21)$

$=48\cdot(-1)+48\cdot(-78)+48\cdot(-21)$

$=48\cdot[(-1)+(-78)+(-21)]$

$=48\cdot(-1-78-21)$

$=48\cdot(-79-21)$

$=48\cdot(-100)$

$=-4800$

7) $29\cdot(-13)+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot13+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot[13+27+(-14)]$

$=-29\cdot(40-14)$

$=-29\cdot26$

$=-754$

$\text{#}Toru$

help

với

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

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

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

41/9:41/18-7<x<(26/5:3.2+45x71/40)

=>-5<x<163/2

=>x ϵ\(\left\{-4;-3;-2;...;81\right\}\).

Ta có:

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

\(=\frac{1}{2}-\frac{7}{12}\)

\(=-\frac{1}{12}.\)

\(\frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)

\(=\frac{1}{48}+\frac{5}{48}\)

\(=\frac{1}{8}.\)

=> \(-\frac{1}{12}< x< \frac{1}{8}\)

=> \(x=0.\)

Vậy \(x=0.\)

Chúc bạn học tốt!

1/2 - ( 4/12 + 3/12 ) < x < 1/48 - ( 6/96 - 16/96 )

1/2 - 7/12 < x < 1/48 + 5/48

-1/12 < x < 1/8

Vì -1/12 < 0

1/8 > 0

nên x = 0

Vậy x = 0

1 + 1/3 + 1/6+1/12+1/24+1/48+1/96
= 2/3 – 1/3 + 1/3 – 1/6+1/6- 1/12+1/12- 1/24+1/24-
1/48+1/48-1/96
=2/3_1/96

= 63/96
= 21/32

chúc cậu hc tốt ạ

1/2 - ( 1/3 + 1/4 ) < ..1....< 1/48 - ( 1/16 - 1/6 )

1

1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

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

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)