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a) 25.6 + 5.5.29 - 45.5 =25.6 - 25.29 + 9.5.5 25.6 - 25.29 +9.(5.5) [có chung 25 nên sẽ đặt 25 ra ngoài rồi trừ cộng trong ngoặc] b) 30.75 + 25.30 - 150 =30.75 + 25.30 - 30.5 [tương tự như trên]
a, 80 - 4 . 5 2 - 3 . 2 3
= 80 – (4.25 – 3.8)
= 80 – 76 = 4
b, 5 6 : 5 4 + 2 3 . 2 2 - 1 2018
= 5 2 + 2 5 - 1
= 25 + 32 – 1 = 56
c, [36.4 – 4. 82 - 7 . 11 2 ]:4 – 2019 0
= (144 – 4. 5 2 ):4 – 1
= (144 – 100):4 – 1
= 11 – 1 = 10
d, 303 – 3.{[655 – (18:2+1). 4 3 +5]}: 10 0
= 303 – 3.(655 – 10.64 + 5):1
= 303 – 10 = 293
a) 80-(4.52-3.23)=80-100+24=4
b)[36.4-4.(82-7.11)2]:4-20190
={4.[36-(82-7.11)2]}:4-1
=[36-(82-7.11)2]-1
=11-1=10
c)56:54+23.22-12018
=52+25-1
=25+32-1=56
d)303-3.{[655-(18:2+1).43+55]}:100
=303-3.[(655-9-1).43+55]:1
=303-3[655-640+5]
=303-3(20)
=303-60=243
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
a) 80- (4.52 - 3.23)
= 80- ( 208 -69 )
=80+139 quy tắc đổi dấu trừ tước dấu ngoặc
= 219
e: Ta có: \(2448:\left[119-\left(23-6\right)\right]\)
\(=2448:\left(119-23+6\right)\)
\(=2448:102=24\)
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a)15-25.8:(100.2) b)80-(4.5\(^2\)-3.2\(^3\)) c)3\(^2\).[(5\(^2\)-3):11]-2\(^4\)+2.10\(^3\) d)2.[(7-3\(^3\):3\(^2\)):2\(^2\)+99]-100
=15-200:(100.2) =80-(4.25-3.8) =9.[(25-3):11]-16+2.1000 =2.[(7.27:9):4+99]-100
=(-185):(100.2) =80-76 =9.[22:11]-16+2.1000 =2.[21:4+99]-100
=(-185):200 =4 =9.2-16+2.1000 =2.104,25-100
=-0,925 =18-16+2.1000 =208,5-100
=18-16+2000 =180,5
=2002
e)303-3.{[655-(18:2+1).4\(^3\)+5]}:10\(^0\) f)(5\(^{2001}\)-5\(^{2000}\)):5\(^{2000}\)
=303-3.{[655-(18:2+1).64+5]}:1 =5\(^{2001}\)-5\(^{2000}\):5\(^{2000}\)
=303-3{[655-10.64+5]}:1 =5\(^{2001}\)-(5\(^{2000}\):5\(^{2000}\))
=303-3{[655-640+5]}:1 =5\(^{2001}\)-5\(^1\)
=303-3.15:1 =5\(^{2000}\)
=303-45:1
=258:1=258
Hok tốt!