Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
a: \(3^8:3^4+2^2\cdot2^3\)
=81+32
=123
b: \(3\cdot4^2-2\cdot3^2\)
\(=48-18\)
=30
a, 38: 34+ 22. 23
= 38-4 + 22+3
= 34 + 25
= 81 + 32
= 113
b, 3 . 42- 2 . 32
= 3 . 16 - 2 . 9
= 48 - 18
= 30
c, 84 : 4 + 39: 37+ 50
= 84 : 4 + 32 + 1
= 84 : 4 + 9 + 1
= 21 + 9 + 1
= 31
d, 295 - ( 31 - 22 . 5)2
= 295 - ( 31 - 4 . 5 )2
= 295 - ( 31 - 20 )2
= 295 - 112
= 295 - 121
= 174
e, 500 - {5[409 - (23 . 3 - 21)2 ] + 103 } : 15
= 500 - {5[409 - (8 . 3 - 21)2 ] + 103 } : 15
= 500 - {5[409 - (24 - 21)2 ] + 103 } : 15
= 500 - {5[409 - 32 ]+ 103 } : 15
= 500 - {5[409 - 9 ]+ 103 } : 15
= 500 - {5 . 400 + 1000 } : 15
= 500 - {2000 + 1000} : 15
= 500 - 3000 : 15
= 500 - 200
= 300
g, 53 . 2 - 100 : 4 + 23 . 5
= 125 . 2 - 100 : 4 + 8 . 5
= 250 - 25 + 40
= 225 + 40
= 265
h, 205 - [1200 - (42 - 2 . 3)3 ] : 40
= 205 - [ 1200 - ( 16 - 2 . 3 )3 : 40
= 205 - [ 1200 - ( 16 - 6 )3 ] : 40
= 205 - [ 1200 - 103 ] : 40
= 205 - [ 1200 - 1000 ] : 40
= 205 - 200 : 40
= 205 - 5
= 200
Đây nha bạn!!!
B=1-2-3+4+5-6-7+8+..........+21-22-23+24
B=(1-2-3+4)+(5-6-7+8)+.......+(21-22-23+24)
B=0+0+............+0
B=0
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)