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a)
\(-68+\left(-72\right)+2017+\left(-28\right)+168.\)
\(=\left(-68+168\right)+\left(-28-72\right)+2017.\)
\(=\left(168-68\right)+\text{ }\left[-\left(72+28\right)\right]+2017.\)
\(=100+\left(-100\right)+2017.\)
\(=100-100+2017.\)
\(=2017\)
b)
\(\left|-28\right|+\left(129-182+99\right)\)\(-\left(129+199-182\right)\)
\(=28+129-182+99\)\(-129-199+182.\)
\(=28+\left(129-129\right)+\left(182-182\right)\)\(-\left(199-99\right)\)
\(=28+0+0-100.\)
\(=-\left(100-28\right)\)
\(=-72\)
c)
\(17\left(-83\right)+18\times83.\)
\(=-17\left(83\right)+18\left(83\right)\)
\(=83\left[18+\left(-17\right)\right]\)
\(=83\left(18-17\right)\)
\(=83\left(1\right)\)
\(=83\)
d)
\(2018^0-\left(15^2\div\left[175+\left(2^3\times5^2-6\times25\right)\right]\right).\text{ }\)
\(=1-\left(15^2\div\left[175+\left(8\times25-6\times25\right)\right]\right)..\)
\(=1-\left(15^2\div\left[175+\left(25\left[8-6\right]\right)\right]\right)\)
\(=1-\frac{15^2}{175+\left(25\times2\right)}.\)
\(=1-\frac{15^2}{175+50}\)
\(=1-\frac{225}{225}\)
\(=1-1\)
\(=0\)
Học tốt
a. 3 . 62 . 8 + 2 .17 . 12 + 4 . 6 . 21
= 186 . 8 + 34 . 12 + 24 . 21
= 1488 + 408 + 504
= 1896 + 504
=2400
a, 32 . 5 - 2x . 6 + 183
= 9 . 5 - 2x . 6 + 1
= 45 - 2x . 6 + 1
= 45 - 2x + 1 . 3 + 1
= 3( 15 - 2x + 1) + 1
b, 100 : { 280 : [ 450 - ( 480 - 4 . 52 ) ] }
= 100 : { 280 : [ 450 - 380 ] }
= 100 : { 280 : 70 }
= 100 : 4 = 25
c, 8 . { 24 - [ 3 . ( 5 + 25 ) + 15 ] : 15 }
= 8 . { 24 - [ 3 . 30 + 15 ] : 15 }
= 8 . { 24 - 105 : 15 }
= 8 . { 24 - 7 }
= 8 . 17 = 136
\(a.\) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\frac{2^3.3^3+3^3.2^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\frac{3^3.\left(-1\right)}{1}=-27\)
\(b.\)\(A=2^2+4^2+6^2+...+20^2=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=2^2.\frac{10.\left(10+1\right).\left(2.10+1\right)}{6}=4.385=1540\)
( Ta có: công thức tính tổng bình phương liên tiếp tứ 1 đến n là: \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\))
\(c.\)\(B=100^2+200^2+...+1000^2=\left(100.1\right)^2+\left(100.2\right)^2+...+\left(100.10\right)^2\)
\(B=100^2.1^2+100^2.2^2+...+100^2.10^2=100^2.\left(1^2+2^2+...+10^2\right)\)
Áp dụng công thức \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Ta có: \(B=100^2\times385=3,850,000\)
a. (83+234)-(34-17)
= (243-34)-(83+17)
= 200-100
= 100
B.(-15)+13+15.(36-25)
= (-15)+15.13+11
= 13+11
= 24