Tính hợp lí
a, 61.49+27.49+12.49+100.51
b, 2 3 . 3 3 . 5 + 9 . 6 3 - 8 . 6 3 . 2018 0 (để kết quả dưới dạng lũy thừa)
c, 25.5.4.27.2
d, 12:{400:[500 – (125+25.7)]}
e, 3 2 . 4 2 . 7 + 12 2 . 6 - 6 2 . 2 2 . 13 + 2020 0
a) 27.49+61.49+49.12+100.51
b) ( 7200+36 ) : 36
a)=49x(27+61+12)+100x51
=49x100+5100
=4900+5100
=10000
b)=7236:36
=201
k nha p
a, 27 . 49 + 61 . 49 + 49 . 12 + 100 . 51
= 27 . 49 + 61 .49 + 12 . 49 + 100 . 51
= ( 27 + 61 + 12 ) . 49 + 100 . 51
= 100 . 49 + 100 . 51
= 100 . ( 49 + 51 )
= 100 . 100
= 10000
b, ( 7200 + 36 ) : 36
= 7200 + 36 : 36 + 1
= 7200 + 1 + 1
= 7202
Tính bằng cách hợp lí
a. 12 × 3 + 4 × 12 + 36
b. 175 - 25 × 2 - 3 × 25 - 50
a)\(=12\times3+4\times12+12\times3=12\times\left(3+4+3\right)=12\times10=120\)
b)\(=25\times7-25\times2-3\times25-2\times25=25\times\left(7-2-3-2\right)=0\)
A) 12x3+4x12+36
=12x3+4x12+12x3
=12x(3+4+3)
=120
Tính hợp lí
a) 1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022
`1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022`
`= (1 - 2) + (3 - 4) + (5 - 6) +...+ (2021 - 2022)`
`= (-1) + (-1) + (-1) + ... + (-1) ` [có `2022 : 2 = 1011` nhóm]
`= (-1) xx 1011 = -1011`
1,tính hợp lí
a.3/5.(-8/3)-3/5:(-3/2)
b-5/6.-12/7-(-21/15)
c.0,125)650.8102
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
Tính hợp lí
a) 1/2 x 1 x 1/3 x 10 x 7/35 x 3/4
\(\dfrac{1}{2}\cdot1\cdot\dfrac{1}{3}\cdot10\cdot\dfrac{7}{35}\cdot\dfrac{3}{4}\)
\(=\dfrac{1}{2}\cdot10\cdot\dfrac{1}{5}\cdot1\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\)
\(=\dfrac{10}{10}\cdot\dfrac{1}{4}=\dfrac{1}{4}\)
Thực hiện phép tính một cách hợp lí
a, A=\(\dfrac{3}{5}.\dfrac{5}{4}-\dfrac{3}{5}.\dfrac{1}{4}\)
Bài 1:tính hợp lí
a)80-(4x5 mũ 2-3x2 mũ 3)
b)125-2x[56-48:(15-7)]
Xong rùi đó>:)) :D
a) 80-(4x5^2-3x2^3)
= 80-(4x25-3x8)
= 80-(100-24)
= 80-76
= 4
b) 125-2x[56-48:(15-7)]
= 125-2x[56-48:8]
= 125-2x[56-6]
=125-2x50
=125-100
=25.
tính hợp lí
a) 7.(-2)3-12.(-5)+(-17)
b) 1632-37-(-157)-163-1532
c) 47.(-918)+(-53).918
d) (-52).(-281)+(-52).181
a: \(7\cdot\left(-2\right)^3-12\cdot\left(-5\right)+\left(-17\right)\)
\(=7\cdot\left(-8\right)+60-17\)
=-56+43
=-13
b: \(1632-37-\left(-157\right)-163-1532\)
\(=\left(1632-1532\right)-37-163+157\)
=100-200+157
=57
c: \(47\cdot\left(-918\right)+\left(-53\right)\cdot918\)
\(=918\left(-47\right)+\left(-53\right)\cdot918\)
\(=918\cdot\left(-47-53\right)\)
\(=918\left(-100\right)=-91800\)
d: \(\left(-52\right)\cdot\left(-281\right)+\left(-52\right)\cdot181\)
\(=\left(-52\right)\left(-281+181\right)\)
\(=\left(-52\right)\cdot\left(-100\right)=5200\)
a: 7⋅(−2)3−12⋅(−5)+(−17)7⋅(−2)3−12⋅(−5)+(−17)
=7⋅(−8)+60−17=7⋅(−8)+60−17
=-56+43
=-13
b: 1632−37−(−157)−163−15321632−37−(−157)−163−1532
=(1632−1532)−37−163+157=(1632−1532)−37−163+157
=100-200+157
=57
c: 47⋅(−918)+(−53)⋅91847⋅(−918)+(−53)⋅918
=918(−47)+(−53)⋅918=918(−47)+(−53)⋅918
=918⋅(−47−53)=918⋅(−47−53)
=918(−100)=−91800=918(−100)=−91800
d: (−52)⋅(−281)+(−52)⋅181(−52)⋅(−281)+(−52)⋅181
=(−52)(−281+181)=(−52)(−281+181)
=(−52)⋅(−100)=5200=(−52)⋅(−100)=5200
Bài 1:cho giá trị của biểu thức
A= (7-3/4 + 1/3) - (6+ 5/4 - 4/3) - (5-7/4 + 5/3)
hãy tính giá trị biểu thức.
Bài 2:Thực hiện hép tính sau một cách hợp lí
a, 1/3 . -4/5 + 1/3 . -6/5
giúp mình với ạ
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)