PT
1
K
Khách
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PT
1
K
10 tháng 9 2018
\(a,165:11\cdot999+85\cdot999=15\cdot999+85\cdot999=999\left(15+85\right)=999\cdot100=99900\)
\(b,25\cdot4+25:5-4\left(30-5\right)-5=25\cdot4+5-4\cdot25-5=\left(25\cdot4-4\cdot25\right)+\left(5-5\right)=0+0=0\)
2 tháng 10 2021
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
2 tháng 10 2021
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 8 2023
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) \(\dfrac{-15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)
\(=\dfrac{-3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)
\(=\left(\dfrac{-3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{5}{5}-\dfrac{7}{3}\)
\(=1-\dfrac{7}{3}\)
\(=-\dfrac{4}{3}\)
13 tháng 8 2023
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
0
H9
HT.Phong (9A5)
CTVHS
17 tháng 1
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) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
LT
2
H9
HT.Phong (9A5)
CTVHS
13 tháng 8 2023
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) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
13 tháng 8 2023
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
HT
1
LM
Lê Minh Vũ
CTVHS
VIP
9 tháng 8 2023
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) \(2015+\left(8\times15-\left(18-8\right)^2\right)\)
\(=2015+\left(8\times15-\left(10\right)^2\right)\)
\(=2015+\left(8\times15-100\right)\)
\(=2015+\left(120-100\right)\)
\(=2015+20\)
\(=2035\)
a, 165:11.999 + 85.999 = (15.11):11.999 + 85.999 = [(15.11):11].999 + 85.999
= 15.(11:11).999 + 85.999 = 15.999 + 85.999 = (15+85).999 = 99900
b, 25.4 + 25:5 – 4.(30 – 5) – 5 = 25.4 + 25:5 – 4.25 – 5 = (25.4 – 4.25) + (25:5 – 5) = 0 + 0 = 0