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a) `27 .75 + 25.27 -150`
`=27 xx (75 + 25)-150`
`=27 xx 100-150`
`=2700-150`
`=2550`
b) `142-[50-(2^3 .10-2^3 .5)]`
`=142-[50-2^3 .(10-5)]`
`=142-(50-8 .5)`
`=142-(50-40)`
`=142-10`
`=132`
c) `375:{32-[4+(5.3^2-42)]}-14`
`=375:{32-[4+(45-42)]}-14`
`=375:[32-(4+3)]-14`
`=375: 25 -14`
`=15-14`
`=1`
a. 27 . 75 + 25 . 27 - 150
= 27. (75 + 25) - 150
= 27 . 100 - 150
= 2700 - 150
= 2550
b. 142 - [50 - (23.10 - 23.5)]
= 142 - [50 - (8 . 10 - 8 . 5)]
= 142 - {50 - [8 . (10 - 5)]}
= 142 - (50 - 40)
= 142 - 10
= 132
c. 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 : {32 –7} – 14
= 375 : 25 - 14
= 15 - 14
= 1
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
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
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; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)
= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)
= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)
= - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)
= -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)
= \(\dfrac{-61}{30}\)
c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)
= - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)
= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)
= \(\dfrac{2}{21}\)
d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)
= - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)
= - \(\dfrac{21}{77}\) - \(\dfrac{22}{77}\)
= - \(\dfrac{43}{77}\)
a, mk làm kết quả lun.
=12:{400:[500-(125+175)]}
=12:{400:[500-300]}
=12:{400:200}
=12:2
=6
b,
=110-18:3
= 110-6
=104
k mk nha. k xong mk làm tiếp. dài quá nhưng cx dễ
\(a)\)
\(27.75+25.75-150\)
\(=2025+675-150\)
\(=2550\)
\(b)\)
\(142-[50-\left(2^3.10-2^3.5\right)]\)
\(=142-[50-\left(80-40\right)]\)
\(=132\)
\(c)\)
\(375:\left\{32-[4+\left(5.3^2-42\right)]\right\}-14\)
\(=375:\left\{32-[4+\left(45-42\right)]\right\}-14\)
\(=375:\left(32-7\right)-14\)
\(=375:25-14\)
\(=1\)
\(d)\)
\(\left\{210:[16+3\left(6+3.2^2\right)]\right\}-3\)
\(=[210:\left(16+3.18\right)]-3\)
\(=210:70-3\)
\(=0\)
\(e)\)
\(500-\left\{5[409-\left(2^3.3-21\right)^2]-1724\right\}\)
\(=500-\left\{5[409-3^2]-1724\right\}\)
\(=500-\left\{5.400-1724\right\}\)
\(=500-\left\{2000-1724\right\}\)
\(=500-276\)
\(=224\)