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a) 5.2² - 18 : 3
= 5.4 - 18 : 9
= 20 - 2
= 18
b) 17 . 85 + 15 . 17 - 120
= 17 . (85 + 15) - 120
= 17 . 100 - 120
= 1700 - 120
= 1580
c) 2³ . 17 - 2³ . 14
= 8 . 17 - 8 . 14
= 8 . (17 - 14)
= 8 . 3
= 24
a) 5 . 22 – 18 : 32
= 5.4 – 18 : 9
= 20 – 2
= 18
b) 17 . 85 + 15 . 17 – 120
= 17. (85 + 15) – 120
= 17.100 – 120
= 170 – 120
= 50
c) 23 . 17 – 23 . 14
= 23.(17 - 14)
= 23.3
= 8.3
= 24
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
2:
1: =>36x+14x=69+81=150
=>50x=150
=>x=3
2: 3^x=81
=>3^x=3^4
=>x=4
3: 3(2x+1)^2=75
=>(2x+1)^2=25
=>2x+1=5 hoặc 2x+1=-5
=>x=-3 hoặc x=2
1:
1: \(\dfrac{13\cdot17^4+4\cdot17^4}{17^3}-\dfrac{14\cdot3^3-14\cdot3^2}{9}\)
\(=\dfrac{17^4\cdot\left(13+4\right)}{17^3}-\dfrac{14\cdot3^2\left(3-1\right)}{9}\)
\(=17\cdot17-14\cdot2\)
=289-28
=261
2:
\(2^3\cdot5^2-\left[131-\left(23-2^3\right)^2\right]\)
\(=8\cdot25-131+\left(-1\right)^2\)
=69+1
=70
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18