27.36 + 73.99+27.14-49.73
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CM
28 tháng 5 2019
a, A = 27.36+73.99+27.14 - 49.73
= 27(36+14)+73(99 - 49)
= 27.50 + 73.50 = 50(27+73) = 50.100 = 5000
b, B = 4 5 . 10 . 5 6 + 25 5 . 2 8 : 2 8 . 5 4 + 5 7 . 2 5
Ta có: 4 5 = 2 2 5 = 2 2 . 5 = 2 10 ; 25 5 = 5 2 5 = 5 2 . 5 = 5 10
=> 4 5 . 10 . 5 6 + 25 5 . 2 8 = 2 10 . 2 . 5 . 5 6 + 5 10 . 2 8 = 2 11 . 5 7 + 5 10 . 2 8 = 2 8 . 2 3 . 5 7 + 5 7 . 5 3 . 2 8 = 2 8 . 5 7 . 2 3 + 5 3
Ta lại có: 2 8 . 5 4 + 5 7 . 2 5 = 2 5 . 2 3 . 5 4 + 5 4 . 5 3 . 2 5 = 2 5 . 5 4 . 2 3 + 5 3
Suy ra B = 2 8 . 5 7 . 2 3 + 5 3 :[ 2 5 . 5 4 . 2 3 + 5 3 ] = 2 8 . 5 7 : 2 5 . 5 4
= 2 8 : 2 5 . 5 7 : 5 4 = 2 3 . 5 3 = 2 . 5 3 = 10 3 = 1000
2 tháng 9 2017
A = 27.36+73.99+27.14-49.73
A=27(36+14)+73(99-49)
A=27.50+79.50
A=50(27+79)
A=50.100=5000
2 tháng 9 2017
27 . 36 + 73 . 99 + 27 . 14 - 49 . 73 = 27 . ( 36 + 14 ) + 73 . ( 99 - 49 )
= 27 . 50 + 73 . 50
= 50 . ( 73 + 27 )
= 50 . 100
= 5000
CHÚC BẠN HOK GIỎI :))
21 tháng 3 2020
\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}\)
\(=\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{18}{7}\)
\(=\frac{1.5}{3.7}-\frac{7.18}{27.7}\)
\(=\frac{5}{21}-\frac{2}{3}\)
\(=\frac{5}{21}-\frac{14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
Chúc bn học tốt
H9
HT.Phong (9A5)
CTVHS
27 tháng 7 2023
a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)
\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)
\(A=27\cdot50+73\cdot50\)
\(A=50\cdot\left(27+73\right)\)
\(A=50\cdot100\)
\(A=5000\)
b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=2^3\cdot5^3\)
\(B=10^3\)
\(B=1000\)
G
1
24 tháng 6 2022
a: \(=\dfrac{1}{2}\cdot\dfrac{2\cdot21+3\cdot27-5\cdot7}{189}\)
\(=\dfrac{1}{2}\cdot\dfrac{88}{189}=\dfrac{44}{189}\)
b: \(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}=\dfrac{1}{2}+\dfrac{4}{5}=\dfrac{13}{10}\)
c: \(=\dfrac{5}{21}-\dfrac{7}{14}\cdot\dfrac{36}{27}=\dfrac{5}{21}-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{5-14}{21}=\dfrac{-9}{21}=\dfrac{-3}{7}\)
V
1
EG
13 tháng 3 2018
\(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)=\frac{1}{2}\cdot\left(\frac{6}{27}-\frac{5}{27}+\frac{3}{7}\right)=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{1}{27}=\frac{3}{14}+\frac{1}{54}=\frac{44}{189}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}=\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
\(\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{36}{14}=\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{18}{7}=\frac{1}{3}\cdot\frac{5}{7}-\frac{6}{7}=\frac{5}{21}-\frac{14}{21}=\frac{-3}{7}\)
7 tháng 2 2016
a/ -388.22
b/ -900
1/ 1/7=5/35
1/5=7/35
mak 1/7<x/35<1/5<=>5/35<x/35<7/35
=>x=6
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
a) \(227+50+23=\left(227+23\right)+50=250+50=300\)
b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)
c) \(1+2+3+4+5+...+97+98+99+100\)
\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)
\(=101+101+101+...+101\)
\(=101\cdot50\)
\(\Leftrightarrow5050\)
d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)
e) \(50-49+48-47+...+4-3+2-1\)
\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)
\(=1+1+1+1+..+1\)
\(=1\cdot25\)
\(=25\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)
\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)
\(=10000\cdot360\)
\(=3600000\)
g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)
h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)
i) \(13\cdot17-256:16+14:7-2021^0\)
\(=13\cdot17-4^4:4^2+2-1\)
\(=13\cdot17-16+2-1\)
\(=13\cdot17-17\)
\(=17\cdot\left(13-1\right)\)
\(=204\)
j) \(7^2-36:3=49-12=37\)