[24+5^2*4–(32–64:2^3)]:(7+24*8)^2
Những câu hỏi liên quan
a) A=40+3/8+7/8+5/8^3+.....+32/8^5
B=24/8^2+40+5/8^2+...............+40/8^4+5/8^4
b) A=1+1/2+1/3+...................+1/64
B=4
Bài 1 So sánh
A= 40+ 3/8 + 7/8^2 + 5/8^3 + 32/8^5
B= 24/8^2 + 40 + 5/8^2 + 40/8^4 + 5/8^4
Bài 2 So sánh
a, 1.3.5.7...99 và 51/2 . 52/2 .... 100/2
b, A= 1+1/2 + 1/3 + 1/4 + ...... + 1/64 và 4
1. (-12)+(-15)+(-31) | 2. 25+37-59 | 3. (-12)+13-(-60) 4. 24:(-2)+(-5) | 5. 32.(-2)+64 | 6. 18.3+(-50):5 7. (-24)+37+163+(-176) | 8. 125+217+(-625)+529 | 9. 29+132+237+868+76310. (-87)+126+187+74 | 11. 29.87+29.13 | 12. 123.(-29)-23.(-29)13. 31.(-18)+31.(-81)-31 | 14.(-12).47+(-12).52+(-12) | 15. 24.(16-5)-16.(24-5)16. 29.(29-3)-19.(29-13) | 17. -452-(-67+75-452) | 18. -(-171+89+223)-(71-111)+223 19. 21.83-3.7.(-17) | 20. 35-{12-[-14+(-2)]} | 21.5+10+15+20+....+100022. 3-6+9-12+....+195-198
Đọc tiếp
1. (-12)+(-15)+(-31) | 2. 25+37-59 | 3. (-12)+13-(-60)
4. 24:(-2)+(-5) | 5. 32.(-2)+64 | 6. 18.3+(-50):5
7. (-24)+37+163+(-176) | 8. 125+217+(-625)+529 | 9. 29+132+237+868+763
10. (-87)+126+187+74 | 11. 29.87+29.13 | 12. 123.(-29)-23.(-29)
13. 31.(-18)+31.(-81)-31 | 14.(-12).47+(-12).52+(-12) | 15. 24.(16-5)-16.(24-5)
16. 29.(29-3)-19.(29-13) | 17. -452-(-67+75-452) | 18. -(-171+89+223)-(71-111)+223
19. 21.83-3.7.(-17) | 20. 35-{12-[-14+(-2)]} | 21.5+10+15+20+....+1000
22. 3-6+9-12+....+195-198
Tính nhẩm8 x 2 ....... 8 x 3 ....... 8 x 4 ....... 8 x 5 .......16 : 8 ..... 24 : 8 ..... 32 : 8 ..... 40 : 8 .....8 x 6 ....... 8 x 7 ....... 8 x 8 ....... 8 x 9 .......48 : 8 ..... 56 : 8 ..... 64 : 8 ..... 72 : 8 .....
Đọc tiếp
Tính nhẩm
8 x 2 = ....... 8 x 3 = ....... 8 x 4 = ....... 8 x 5 = .......
16 : 8 = ..... 24 : 8 = ..... 32 : 8 = ..... 40 : 8 = .....
8 x 6 = ....... 8 x 7 = ....... 8 x 8 = ....... 8 x 9 = .......
48 : 8 = ..... 56 : 8 = ..... 64 : 8 = ..... 72 : 8 = .....
8 x 2 = 16 8 x 3 = 24 8 x 4 = 32 8 x 5 = 40
16 : 8 = 2 24 : 8 = 3 32 : 8 = 4 40 : 8 = 5
8 x 6 = 48 8 x 7 = 56 8 x 8 = 64 8 x 9 = 72
48 : 8 = 6 56 : 8 = 7 64 : 8 = 8 72 : 8 = 9
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16 24 32 40
2 3 4 5
48 56 64 72
6 7 8 9
Viết sô thích hợp vào chỗ chấm điểm
A) 3/4=3*4/4*4=…\……
2/7=2*.../7*3=…\…
24/42=24:6/42:…=…\…
B) 2/3=6/…
18/24=6/…
2/7=8/…=…\35
3/5=…/20
32/24=…/3
48/72=…/9=2/…=…/24
A) 12/16
2*3; 6/21
42:6; 4/7
B) 6/9
6/8
8/28=10/35
12/20
4/3
6/9=2/3=16/28
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Tính nhẩm
16 : 8 =
24 : 8 =
32 : 8 =
40 : 8 =
16 : 2 =
24 : 3 =
32 : 4 =
40 : 5 =
16 : 8 = 2
24 : 8 = 3
32 : 8 = 4
40 : 8 = 5
16 : 2 = 8
24 : 3 = 8
32 : 4 = 8
40 : 5 = 8.
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tính
a. 8/5+7/6+5/9-2
b.3-5/6-4/9+32/24
a)\(\dfrac{8}{5}+\dfrac{7}{6}+\dfrac{5}{9}-2\)
\(=\dfrac{144}{90}+\dfrac{105}{6}+\dfrac{50}{90}-\dfrac{180}{90}=\dfrac{119}{90}\)
b) \(3-\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{32}{24}\)
\(=3-\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{54}{18}-\dfrac{15}{18}-\dfrac{8}{18}+\dfrac{24}{18}=\dfrac{55}{18}\)
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11 tháng 1 2022 lúc 14:55
Cho A = 40 + \(\dfrac{3}{8}+\dfrac{7}{8^2}+\dfrac{5}{8^3}+\dfrac{32}{8^5}\)
B = \(\dfrac{24}{8^2}+40+\dfrac{5}{8^2}+\dfrac{40}{8^4}+\dfrac{5}{8^4}\)
So sánh A và B
Bài 1. Tính hợp lý1) (–12) +6.(–3) 2) (36 -2020) + (2019 -136) – 273) (144 – 97) – (244 – 197) 4) (–24).13 – 24.( –3)5) 54+55+56+57+58-(64+65+66+67+68) 6) 24(16 – 5) – 16(24 – 5) 7) 47.(23 + 50) – 23.(47 + 50)8) (-31). 47 + (-31). 52 + (-31)Bài 2: Tìm số nguyên x, biết: 1)-17-(2x-5)-62) 10-2(4-3x)-43)-12+3(-x+7)-184)-45:[5.(-3-2x)]35) x.(x+3)06) (x-2).(x+4)07) x.(x+1).(x-3)0
Đọc tiếp
Bài 1. Tính hợp lý
1) (–12) +6.(–3)
2) (36 -2020) + (2019 -136) – 27
3) (144 – 97) – (244 – 197)
4) (–24).13 – 24.( –3)
5) 54+55+56+57+58-(64+65+66+67+68)
6) 24(16 – 5) – 16(24 – 5)
7) 47.(23 + 50) – 23.(47 + 50)
8) (-31). 47 + (-31). 52 + (-31)
Bài 2: Tìm số nguyên x, biết:
1)-17-(2x-5)=-6
2) 10-2(4-3x)=-4
3)-12+3(-x+7)=-18
4)-45:[5.(-3-2x)]=3
5) x.(x+3)=0
6) (x-2).(x+4)=0
7) x.(x+1).(x-3)=0
Bài 1:
1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)
\(=-12-18\)
=-30
2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)
\(=36-2020+2019-136-27\)
\(=1-100-27\)
\(=-126\)
3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)
\(=144-97-244+197\)
\(=-100+100=0\)
4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)
\(=-24\cdot13+24\cdot3\)
\(=24\cdot\left(-13+3\right)\)
\(=24\cdot\left(-10\right)=-240\)
5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)
\(=54+55+56+57+58-64-65-66-67-68\)
\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)
\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)
=-50
6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)
\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)
\(=-24\cdot5+16\cdot5\)
\(=5\cdot\left(-24+16\right)\)
\(=-5\cdot8=-40\)
7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)
\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)
\(=47\cdot50-23\cdot50\)
\(=50\cdot\left(47-23\right)\)
\(=50\cdot24=1200\)
8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)
\(=-31\cdot\left(47+52+1\right)\)
\(=-31\cdot100=-3100\)
Bài 2:
1) Ta có: \(-17-\left(2x-5\right)=-6\)
\(\Leftrightarrow-17-2x+5+6=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow-2x=6\)
hay x=-3
Vậy: x=-3
2) Ta có: \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x+4=0\)
\(\Leftrightarrow6x+6=0\)
\(\Leftrightarrow6x=-6\)
hay x=-1
Vậy: x=-1
3) Ta có: \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow-12-3x+21+18=0\)
\(\Leftrightarrow-3x+27=0\)
\(\Leftrightarrow-3x=-27\)
hay x=9
Vậy: x=9
4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)
\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)
\(\Leftrightarrow-2x-3=-3\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: x=0
5) Ta có: x(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)
6) Ta có: (x-2)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-1;3\right\}\)
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Bài 1:
1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)
=−12−18=−12−18
=-30
2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27
=36−2020+2019−136−27=36−2020+2019−136−27
=1−100−27=1−100−27
=−126
Tớ chcs cậu học thật giỏi nha !