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A = \(\dfrac{72^3.5^{42}}{108^4}\)
= \(\dfrac{\left(2^3.3^2\right)^3.5^{42}}{\left(2^2.3^3\right)^4}\)
= \(\dfrac{2^9.3^6.5^{42}}{2^8.3^{12}}\)
= \(\dfrac{2.5^{42}}{3^6}\)
@trần phúc nguyên
B = \(\dfrac{45^3.20^4.18^2}{180^5}\)
= \(\dfrac{\left(3^2.5\right)^3\left(2^2.5\right)^4.\left(2.3^2\right)^2}{\left(2^2.3^2.5\right)^5}\)
= \(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}\)
= \(\dfrac{3^{10}.5^7.2^{10}}{2^{10}.3^{10}.5^5}\)
= 25
@trần phúc nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
a, [ 58+(-42)]-(58-42)
= 58 - 42 - 58 + 42
= ( 58 - 58 ) + ( 42 - 42)
= 0+0=0
b, ( 65 - 35 ) - ( 35 + 27 )
= 65 - 35 - 35 - 27
= 65 - 70 -27
= 65 - 97
= -32
c, 3 . 52 - 176 : 23
=3 .25 - 176 : 8
= 75 - 22
= 53
d, 22 . 3 - 5 . 23
= 4 . 3 - 5 . 8
= 12 - 40
= -28
g, 2020 - [ 39 - ( 5 - 1 )2 ] - 2019
= 2020 - [ 39 - 42 ] - 2019
= ( 2020 -2019 ) - [ 39 - 16 ]
= 1 - 23
= -22
k, 2018 : [ 38 - ( 7 - 1 )2 ] + 20180
= 2018 : [ 38 - 62 ] + 1
= 2018 : [ 38 - 36 ] + 1
= 2018 : 2 + 1
= 1009 + 1 =1010
a)\(\left[58+\left(-42\right)\right]-\left(58-42\right)\)
=\(58-42-58+42\)
=\(0\)
b)\(\left(65-35\right)-\left(35+27\right)\)
=\(65-35-35+27\)
=\(-32\)
c)\(3.5^2-176:2^3\)
=\(75-22\)
=\(53\)
d)\(2^2.3-5.2^3\)
=\(2^2\left(3-10\right)\)
=\(-28\)
g)\(2020-\left[39-\left(5-1\right)^2\right]-2019\)
=\(1-\left(39-16\right)\)
=\(-22\)
k)\(2018:\left[38-\left(7-1\right)^2\right]+2018^0\)
=\(2018:2+2018^0\)
=\(1010\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2.5^{x+1}+3.5^x=2^2.3.1\)
\(5^x.\left(10+3\right)=13\)
\(5^x=1\)
x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{16}{51}\)
a) 1/1.2+1/2.3+1/3.4+...+1/99.100
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
b) 2/3.5+2/5.7+...+2/49.51
= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )
= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )
= 2 . ( 1/3 - 1/50 )
= 2 . 47/150
= 47/75
(3.5-9³).(1+2.3²)+42
= (15 - 729).(1 + 18) + 42
= (-714).19+ 42
= -13524
\(\left(3.5-9^3\right).\left(1+2.3^2\right)+42\)
\(=\left(15-729\right).\left(1+2.9\right)+42\)
\(=-714.\left(1+18\right)+42\)
\(=-714.19+42\)
\(=-13566+42\)
\(=-13524\)
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