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
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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: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
`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`
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:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
c) C= 1+2-3-4+5+6-7-8+...-111-112+113+114+115
ta thấy : 114 chia 4 dư 2 ; 115 chia 4 dư 3
=> C=1+(2-3-4+5)+(6-7-8+9)+...+((110-111-112+113)+114+115
=> C=1+0+0+...+0+229
=> C=300
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
a: =382-282+531-331
=100+200=300
b: =(7-8)+(9-10)+...+(2009-2010)
=(-1)+(-1)+....+(-1)
=-1002
c: =-(1+2+3+...+2009+2010)
=-2010*2011/2=-2021055
\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)
\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)
\(=2018\times0=0\)
B-a:tương tự bài A
b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{9}\times1=\frac{7}{9}\)\
thanks
B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)
\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)
thanks
a)= 2021.2021-2020.(2021+1)
= 2021.(2020+1)-2020.(2021+1)
= (2021.2020)+2021-(2020.2021)-2020
= 1
b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
B= -4+(-4)+....................(-4)+2021
B= -4x505+2021
B= -2020 + 2021
B = 1
Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
a, 61.49+27.49+12.49+100.51 = 49.(61+27+12)+100.51 = 49.100+51.100 = 100.(49+51) = 100.100 = 10000
b, 2 3 . 3 3 . 5 + 9 . 6 3 - 8 . 6 3 . 2018 0 (để kết quả dưới dạng lũy thừa)
= 6 3 . 5 + 6 3 . 9 - 8 . 6 3 . 1 = 6 3 . ( 5 + 9 - 8 ) = 6 4
c, 25.5.4.27.2 = (25.4).(5.2).27 = 100.10.27 = 27000
d, 12:{400:[500 – (125+25.7)]} = 12:{400:[500 – 300]} = 12:(400:200) = 12:2 = 6
e, 3 2 . 4 2 . 7 + 12 2 . 6 - 6 2 . 2 2 . 13 + 2020 0
= 12 2 . 7 + 12 2 . 6 - 12 2 . 3 + 1
= 12 2 . 7 + 6 - 13 + 1 = 1