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1: \(23+\left(-13\right)+\left(-50\right)\)
\(=23-13-50\)
=10-50
=-40
2: \(-5+15+\left(-123\right)\)
\(=\left(-5+15\right)-123\)
=10-123
=-113
3: \(5871:\left\{928-\left[\left(-82+247\right)\right]\right\}\cdot5\)
\(=5871-\left\{928+82-247\right\}\cdot5\)
\(=5871-763\cdot5=5871-3815=2056\)
4: \(40-\left(4\cdot5^2-3\cdot2^3\right)\)
\(=40-4\cdot5^2+3\cdot2^3\)
\(=40-4\cdot25+3\cdot8\)
=40-100+24
=64-100
=-36
5: \(6^2\cdot5-7^2+149\)
\(=36\cdot5-49+149\)
\(=180+149-49\)
=180+100
=280
6: \(-210:\left[16+3\cdot\left(6+3\cdot2^2\right)\right]+\left(-2022\right)\)
\(=-210:\left[16+3\cdot\left(6+3\cdot4\right)\right]+\left(-2022\right)\)
\(=-210:\left[16+3\cdot18\right]+\left(-2022\right)\)
\(=-210:70-2022\)
=-3-2022
=-2025
7: \(5\cdot2^3+7^{11}:7^9-2023^0\cdot1^8\)
\(=5\cdot8+7^2-1\)
=40+49-1
=39+49
=88
8: \(400:\left\{5\cdot\left[360-\left(290+2\cdot5^2\right)\right]\right\}\)
\(=400:\left\{5\cdot\left[360-290-2\cdot25\right]\right\}\)
\(=400:\left\{5\cdot20\right\}\)
\(=\dfrac{400}{100}=4\)
9: \(75-\left(3\cdot5^2\right)-4\cdot5^3\)
\(=75-3\cdot25-4\cdot5^3\)
=-4*125
=-500
\(\left\{210:\left[16+3.\left(6+3.22\right)\right]\right\}-3\)
\(=\left\{210:\left[16+3.\left(6+66\right)\right]\right\}-3\)
\(=\left\{210:\left[16+3.72\right]\right\}-3\)
\(=\left\{210:\left[16+216\right]\right\}-3\)
\(=\left\{210:232\right\}-3\)
\(=\dfrac{105}{116}-3\)
\(=\dfrac{-243}{116}\)
Bài 1: Thực hiện các phép tính sau:
\(a)\)Chưa rỏ đề
\(b)\)\(5025\div5-25\div5\)
\(=\)\(1005-5\)
\(=\)\(1000\)
\(c)\)\(218-180\div2\div9\)
\(=\)\(218-10\)
\(=\)\(208\)
\(d)\)\(\left(328-8\right)\div32\)
\(=\)\(320\div32\)
\(=\)\(10\)
Bài 1:
a) ( Tôi không nhìn rõ đầu bài )
b) 5025 : 5 - 25 : 5
= ( 5025 - 25 ) : 5
= 5000 : 5
= 1000
c) 218 - 180 : 2 : 9
= 218 - 180 : ( 2 . 9 )
= 218 - 180 : 18
= 218 - 10
= 208
d) ( 328 - 8 ) : 32
= 320 : 32
= 10
a, 27.75 + 25.27 – 150
= 2025 + 675 – 150 = 2550
b, 142 – [50 – ( 2 3 .10 – 2 3 .5)]
= 142 – [50 – (80 – 40)] = 132
c, 375:{32 – [4+( 5 . 3 2 – 42]} – 14
= 375:{32 – [4+(45 – 42)]} – 14
= 375:(32 – 7) – 14 = 15 – 14 = 1
d, {210:[16+3.(6+3. 2 2 )]} – 3
= [210:(16+3.18)] – 3
= 210 : 70 – 3 = 3 – 3 = 0
a, 27.75 + 25.27 – 150
= 2025 + 675 – 150 = 2550
b, 142 – [50 – ( 2 3 .10 – 2 3 .5)]
= 142 – [50 – (80 – 40)] = 132
c, 375:{32 – [4+( 5 . 3 2 – 42]} – 14
= 375:{32 – [4+(45 – 42)]} – 14
= 375:(32 – 7) – 14 = 15 – 14 = 1
d, {210:[16+3.(6+3. 2 2 )]} – 3
= [210:(16+3.18)] – 3
= 210 : 70 – 3 = 3 – 3 = 0
\(a,\dfrac{5}{16}+\dfrac{-5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\\ b,\dfrac{-2}{7}+\dfrac{3}{-5}=\dfrac{-10}{35}-\dfrac{21}{35}=\dfrac{-10-21}{35}=\dfrac{-31}{35}\\ c,\dfrac{-2}{-5}+\dfrac{-5}{-6}=\dfrac{2}{5}+\dfrac{5}{6}=\dfrac{12}{30}+\dfrac{25}{30}=\dfrac{12+25}{30}=\dfrac{37}{30}\\ d,\dfrac{-5}{-3}+\dfrac{3}{-7}=\dfrac{5}{3}-\dfrac{3}{7}=\dfrac{35}{21}-\dfrac{9}{21}=\dfrac{35-9}{21}=\dfrac{26}{21}\)
a)Có.Bởi vì các chữ số trong phép tính chia hết cho 7
b)Không.Do 35 - 25 không chia hết cho 7
c)Có.Do 16 + 40 chia hết cho 7 và 490 chia hết cho 7
a) Ta có : \(\left\{{}\begin{matrix}28⋮7\\42⋮7\\210⋮7\end{matrix}\right.\Rightarrow28+42+210⋮7\)
b 35 chia hết cho 7, -25 không chia hết cho 7, 140 chia hết cho 7
=> 35-25+140 không chia hết cho 7
c) 16 chia 7 dư 2, 40 chia 7 dư 5 nên 16+40 chia hết cho 7
490 chia hết cho 7
=> 16+40+490 chia hết cho 7
a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)
\(=142-\left[50-80+40\right]\)
=142-10
=132
a. 142 - [50-(23.10-23.5)]
= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]
= 142 - [ 50 - ( 80 - 40 )]
= 142 - [ 50 - 40 ]
= 142 - 10 = 132
b. 375 : { 32[ 4+(5.32 - 42 )]} - 14
= 375 : { 32[4+(5.9 - 42 )]} - 14
= 375 : { 32[4 + ( 45 - 42 )]} - 14
= 375 : {32[4+3]} - 14
= 375 : 224 - 14
c.{210 : [ 16+3.(6+3.2^2)]} - 3
= {210 : [ 16 + 3.(6+3.4)]} - 3
= { 210 :[16+3.(6+12)]}-3
= {210 : [ 16+3.18)]} - 3
= { 210 : [ 16 + 54]} - 3
= { 210 : 70 } - 3
= 30 - 3 = 27
d.500-{5.[409-(2^3 . 3-21^2)] - 1724}
= 500-{5.[409-(8 . 63 - 21^2] - 1724}
= 500 - { 5.[409- (504 - 441) - 1724}
= 500 -{ 5.[ 409 - 63 ] - 1724}
= 500 - { 5.346 - 1724}
= 500 - { 1720 - 1724 }
= 500 - 6
= 494
\(\left\{210:\left[16+3.\left(6+3.2^2\right)\right]\right\}-3\)
\(=\left\{210:\left[16+3.\left(6+3.4\right)\right]\right\}-3\)
\(=\left\{210:\left[16+3.\left(6+12\right)\right]\right\}-3\)
\(=\left\{210:\left[16+3.18\right]\right\}-3\)
\(=\left\{210:\left[16+54\right]\right\}-3\)
\(=\left\{210:70\right\}-3\)
\(=3-3\)
\(=0\)