a) \(\left|5\right|+\left|-5\right|=5+5=10\)

b) \(\left|-25\right|-\left|-20\right|=25-20=5\)

c) \(\left|10\right|.\left|-16\right|=10.16=160\)

d) \(\left|-49\right|:\left|-7\right|=49:7=7\)

a ) l 5 l + l - 5 l = 5 + 5 = 10

b ) l - 25 l - l - 20 l = 25 - 20 = 5

c ) l 10 l . l - 16 l = 10 . 16 = 160

d ) l - 49 l : l - 7 l = 49 : 7 = 7

a, \(\left|5\right|+\left|-5\right|=5+5=10\)

b, \(\left|-25\right|-\left|-20\right|=25-20=5\)

c \(\left|10\right|.\left|-16\right|=10.16=160\)

d, \(\left|-49\right|:\left|7\right|=49:7=7\)

a) |5| + |-5| = 5 + 5 = 10

b) |-25| - |-20| = 25 - 20 = 5

c) |10| . |-16| = 10 . 16 = 160

d) |-49| : |7| = 49 : 7 = 7

A

\(\dfrac{25}{49}\) a nhé

A

1/2×(2×x-3)+105/2=-137/2

1/2×(2×x-3)=-137/2-105/2

1/2×(2×x-3)=-242/2

2×x-3=-242/2:1/2

2×x-3=-242

2.x=(-242)+3

2.x=239

x=239:2

x=239/2

A,-12×25-25×75-25×13

=[(-12)-75-13].25

=(-100).25

-2500

B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5

=(-35)-(-25)+(-13)

=-23

C,-354+265-156+125

=[(-354)-156]+(265+125)

=(-510)+277

=-233

D,-74+40-50+16-35

=(-74)+40+(-50)+16+(-35)

=[(-74)+(-35)]+[40+(-50)+16]

=(-109)+26

=-83

E,-2/3×4/5+-4/5×4/3-0,125

=-2/3.4/5+4/5.(-4/3)-0,125

=4/5.[-2/3+(-4/3)]-0,125

=4/5.(-2)-0,125

=-8/5-0,125

=(-1,6)+(-0,125)

=-1,725

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

=  1 + 1 + 8

=  2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)

=  \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))

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

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8

= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8

= \(\dfrac{1}{2}\) + 4

= \(\dfrac{9}{2}\) 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

  = (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

= 1 + 1 + 8

= 2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)

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

= \(\dfrac{1}{2}\) x 10

= 5

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

|5|+|-5|=5+5=10

|10|x|-16|=10x16=160

|-25|-|-20|=25-20=5

|-49|:|7|=49:7=7