1+1-1-1+2-3+1
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Tính:
A=(1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4)...(1-1/1+2+3+4+...+2022)
B=1+1/2(1+2)+1/3(1+2+3)+1/100(1+2+3+...+100)
Tính:
1 + 2 = 1 + 1 = 1 + 2 = 1 + 1 + 1 =
1 + 3 = 2 - 1 = 3 - 1 = 3 - 1 - 1 =
1 + 4 = 2 + 1 = 3 - 2 = 3 - 1 + 1 =
- Nhẩm tính rồi điền kết quả vào chỗ trống.
- Biểu thức có hai phép tính thì thực hiện từ trái sang phải.
1 + 2 = 3 1 + 1 = 2 1 + 2 = 3 1 + 1 + 1 = 3
1 + 3 = 4 2 - 1 = 1 3 - 1 = 2 3 - 1 - 1 = 1
1 + 4 = 5 2 + 1 = 3 3 - 2 = 1 3 - 1 + 1 = 3
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tính;
1+2=3 1+1=2 1+2=3 1+1+1=3
1+3=4 2-1=1 3-1 =2 3-1-1= 0
1+4=5 2+1=3 3-2=1 3-1+1=3
Tính tổng sau: a) 1/2+1/6+1/12+1/20+1/30 b) 1/15+1/35+1/63+1/99+1/143 c) 1/6+1/12+1/20+1/30+1/42+1/56 d) 1/2+1/2^2+1/2^3+1/2^4+1/2^5 e) 1/7+1/7^2+1/7^3+...+1/7^100 f) 1+1/2*(1+2)+1/3*(1+2+3)+1/4*(1+2+3+4)+...+1/200*(1+2+3+..+200) g) (1/2+1)*(1/3+1)*(1/4+1)*..*(1/100+1) h) (1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/2022) Giúp mk vs ạkkk
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
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Tính?
1 + 1 …
1 + 2 …
2 + 2 …
1 + 1 …
2 + 1 …
1 + 3 …
3 + 1 …
1 + 2 …
3 + 1 …
1 + 1 …
1 + 3 …
2 + 1 …
Đọc tiếp
Tính?
| 1 + 1 =… | 1 + 2 =… | 2 + 2 = … | 1 + 1 =… |
| 2 + 1 =… | 1 + 3 =… | 3 + 1 =… | 1 + 2 =… |
| 3 + 1 =… | 1 + 1 =… | 1 + 3 =… | 2 + 1 =… |
Lời giải chi tiết:
| 1 + 1 = 2 | 1 + 2 = 3 | 2 + 2 = 4 | 1 + 1 = 2 |
| 2 + 1 = 3 | 1 + 3 = 4 | 3 + 1 = 4 | 1 + 2 = 3 |
| 3 + 1 = | 1 + 1 = 2 | 1 + 3 = 4 | 2 + 1 = 3 |
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Tính:
2 - 1 = 3 - 1 = 1 + 1 = 1 + 2 =
3- 1 = 3 - 2 = 2 - 1 = 3 - 2 =
3 -2 = 2 - 1 = 3 - 1 = 3 - 1 =
2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
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2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
ok nhá
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Tính:
1 + 2 …
3 – 1 …
1 + 1 …
2 – 1 …
3 – 2 …
3 – 2 …
2 – 1 …
3 – 1 …
3 – 1 …
2 – 1 …
3 – 1 …
3 – 2 …
Đọc tiếp
Tính:
| 1 + 2 = … | 3 – 1 = … | 1 + 1 = … | 2 – 1 = … |
| 3 – 2 = … | 3 – 2 = … | 2 – 1 = … | 3 – 1 = … |
| 3 – 1 = … | 2 – 1 = … | 3 – 1 = … | 3 – 2 = … |
Lời giải chi tiết:
| 1 + 2 = 3 | 3 – 1 = 2 | 1 + 1 = 2 | 2 – 1 = 1 |
| 3 – 2 = 1 | 3 – 2 = 1 | 2 – 1 = 1 | 3 – 1 = 2 |
| 3 – 1 = 2 | 2 – 1 = 1 | 3 – 1 = 2 | 3 – 2 = 1 |
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| 1+2=3 | 3-1=2 | 1+1=2 | 2-1=1 |
| 3-2=1 | 3-2=1 | 2-1=1 | 3-1=2 |
| 3-1=2 | 2-1=1 | 3-1=2 | 3-2=1 |
#HT#
1. 3, 2, 2, 1.
2. 1 ,1 ,1 ,2.
3. 2, 1, 2, 1.
HT!
Xem thêm câu trả lời
B=1/3+1/3^2+1/3^3+1/3^4)*3^5+(1/3^5+1/3^6+1/3^7+1/3^8)*3^9+.....+(1/3^97+1/3^98+1/3^99+1/3^100)*3^101
C=(1/4-1)*(1/9-1)*(1/16-1)*..........*(1/81-1)*(1/100-1)
D=(1/2-1)(1/2+1)(1/2^2+1)(1/2^4+1)(1/2^8+1)(1/2^16+1)
Bạn nào làm đúng cả 3 câu này mình tick cho 3 cái
so sánh
a)A=1/2^1+1/2^2+1/2^3+...+1/2^49+1/2^50 với 1
b)B=1/3^1 +1/3^2+1/3^3...+1/3^99+1/3^100 với 1/2
c)C=1/4^1+1/4^2+1/4^3+...+1/4^999+1/4^1000 với 1/3
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}
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Bạn Detective_conan giải đúng đấy!
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Tính:
| 1 + 2 = … | 1 + 1 = … | 1 + 1 + 1 = … |
| 3 – 1 = … | 2 – 1 = … | 3 – 1 – 1 = … |
| 3 – 2 = … | 2 + 1 = … | 3 – 1 + 1 = … |
Lời giải chi tiết:
| 1 + 2 = 3 | 1 + 1 = 2 | 1 + 1 + 1 = 3 |
| 3 – 1 = 2 | 2 – 1 = 1 | 3 – 1 – 1 = 1 |
| 3 – 2 = 1 | 2 + 1 = 3 | 3 – 1 + 1 = 3 |
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A=(1 - 1/1+2) x (1-1/1+2+3) x (1-1/1+2+3) x (1-1/1+2+3)x1-1/1+2+3+4)x ... x (1-1/1+2+3+...+2018)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+4+...+2018}\right)\)
\(A=\frac{2}{1+2}\cdot\frac{2+3}{1+2+3}\cdot\frac{2+3+4}{1+2+3+4}\cdot...\cdot\frac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018}\)
Đến chỗ này đố ai tính được ?!!?!