tính nhanh
(1+2+3+4+...+2019) x (300:4-75)
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( 1 + 3 + 5 + 7 +...+ 2019) \(\times\) ( 4848 \(\times\) 75 - 7575 \(\times\) 48):51
=(1 + 3 + 5 + 7 +...+2019)\(\times\)( 48 \(\times\) 101 \(\times\) 75 - 75 \(\times\) 101 \(\times\)48):51
=(1+3+5+7+...+2019) \(\times\) ( 48 \(\times\) 101 \(\times\) 75 - 48 \(\times\) 101 \(\times\) 75):51
=(1+3+5+7...+2019)\(\times\)0 : 51
= 0: 51
= 0
1, => [ 462 + 642 - 2x ] : 2 = 531
1104 -2x = 1062
2x = 42
x =21
Bài 1:
a) =25,97+(6,54+103,46)
=25,97+110
=135,97
b)136x75+75x64
=75x(136+64)
=75x200
=15 000
c) (21/8+1/2):5/16
=(21/8+4/8)x16/5
=25/8x16/5
=10
d)3/17-4/5+14/17
=(3/17+14/17)-4/5
=1-4/5
=1/5
Bài 2:
a)720:\([41-(2x-5)]\)=120
41 - (2x-5) =720:120
41 - (2x-5) =6
2x-5 =41-6
2x-5 =35
2x =35+5
2x =40
x =40:2
x =20
b)2/3 x X +3/4=3
2/3 x X =3-3/4
2/3 x X =12/4-3/4
2/3 x X =9/4
x =9/4:2/3
x =9/4x3/2
x =27/8
c) x+0,34=1,19x1,02
x+0,34=1,2138
x =1,2138-0,34
x =0,8738
\(Q=\)\(1-2+3-4+...+2017-2018+2019\)
\(Q=\left(1-2\right)+\left(3-4\right)+...+\left(2017-2018\right)+2019\)
\(Q=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2019\)
\(Q=\left(-1\right).1009+2019\)
\(Q=\left(-1009\right)+2019\)
\(Q=1010\)
~~~~~~~~~~~~~~Hok tốt~~~~~~~~~~~~~~~~~
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
(1+3+5+7+...+2019+2021)
A=1−3+5−7+......−2019+2021−2023
A=(1−3)+(5−7)+....+(2021−2023)A=(1−3)+(5−7)+....+(2021−2023)
A=−2+(−2)+....+(−2)(506)A=−2+(−2)+....+(−2)(506cặp)
a=−2.506A=−2.506
A=−1012A=−1012