\(a,165:11\cdot999+85\cdot999=15\cdot999+85\cdot999=999\left(15+85\right)=999\cdot100=99900\)

\(b,25\cdot4+25:5-4\left(30-5\right)-5=25\cdot4+5-4\cdot25-5=\left(25\cdot4-4\cdot25\right)+\left(5-5\right)=0+0=0\)

Cảm ơn.

Bài 3:

a: Ta có: \(23\left(42-x\right)=23\)

\(\Leftrightarrow42-x=1\)

hay x=41

b: Ta có: 15(x-3)=30

nên x-3=2

hay x=5

Bài 1: 

a: 32+89+68=100+89=189

b: 64+112+236=300+112=412

c: \(1350+360+650+40=2000+400=2400\)

a) \(\dfrac{-5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{2}{9}\right)-\left(\dfrac{4}{15}+\dfrac{11}{15}\right)\)

\(=\dfrac{-3}{9}-\dfrac{15}{15}\)

\(=-\dfrac{1}{3}-1\)

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

b) \(\dfrac{-8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{6}\)

\(=\left(\dfrac{-8}{13}+\dfrac{21}{13}\right)-\dfrac{7}{16}-\dfrac{1}{6}\)

\(=\dfrac{13}{13}-\dfrac{29}{48}\)

\(=1-\dfrac{29}{48}\)

\(=\dfrac{19}{48}\)

c) \(\dfrac{-15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)

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

\(=\left(\dfrac{-3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)

\(=\dfrac{5}{5}-\dfrac{7}{3}\)

\(=1-\dfrac{7}{3}\)

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

a: =-5/9+2/9-11/15-4/15

=-3/9-1

=-12/9=-4/3

b: =-8/13+21/13-7/16-1/6

=1-1/6-7/16

=5/6-7/16

=19/48

c: =-15/25+8/5-2/3-5/3

=-7/3+1

=-4/3

Ta có: \(S=\frac{1}{5.6}+\frac{2}{6.8}+\frac{3}{8.11}+\frac{4}{11.15}+\frac{5}{15.20}\)

\(\Rightarrow S=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{15}-\frac{1}{20}\)

\(=\frac{1}{5}-\frac{1}{20}=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)

a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)

\(=1-\dfrac{29}{15}\)

\(=\dfrac{15-29}{15}\)

\(=-\dfrac{14}{15}\) 

b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)

\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=\dfrac{7}{9}\cdot1\)

\(=\dfrac{7}{9}\)

c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\) 

\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)

\(=\dfrac{25}{63}\)

d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)

\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)

\(=3\cdot\dfrac{9}{11}\)

\(=\dfrac{27}{11}\)

a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)

\(=\dfrac{2}{5}+\dfrac{4}{2}\)

\(=\dfrac{2}{5}+2\)

\(=\dfrac{2}{5}+\dfrac{10}{5}\)

\(=\dfrac{12}{5}\)

b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)

\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)

\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)

\(=1-\dfrac{2}{2008}\)

\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)

\(=\dfrac{2006}{2008}\)

\(=\dfrac{1003}{1004}\)

a: =2/5+4/2

=2/5+2

=12/5

b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

a) \(18+25\times4-4^3\)

\(=18+100-64\)

\(=118-64\)

\(=54\)

b) \(275-\left(49+125\div5^3\right)\)

\(=275-\left(49+125\div125\right)\)

\(=275-\left(49+1\right)\)

\(=275-49-1\)

\(=226-1\)

\(=225\)

c) \(2015+\left(8\times15-\left(18-8\right)^2\right)\)

\(=2015+\left(8\times15-\left(10\right)^2\right)\)

\(=2015+\left(8\times15-100\right)\)

\(=2015+\left(120-100\right)\)

\(=2015+20\)

\(=2035\)