RGBT a,A= (1 1/2)(1 1/4)(1 1/16)...(1 1/2^2n) ...

MT

Mai Trung Quốc

11 tháng 8 2020 - olm

RGBT a,A= (1+1/2)(1+1/4)(1+1/16)...(1+1/2^2n) b,B= (10+1)(10^2+1)...(10^2n+1) làm nhanh nha bạn và sang kênh youtube ủng hộ cho mình nha bạn! ...

Đọc tiếp

#Toán lớp 8

0

NT

nguyen thua tuan

17 tháng 7 2017 - olm

RGBT:

(1+1/2)(1+1/4)(1+1/16)..........(1+1/2²ⁿ)

Chú ý:bài này thuộc phần hiệu 2 bình phương.

#Toán lớp 8

0

TM

Tô Mì

26 tháng 5 2022

Tính \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2n}\)

#Toán lớp 8

2

TC

TV Cuber

26 tháng 5 2022

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{16}-....-\dfrac{1}{2n}\)

\(A=1-\dfrac{1}{2n}\)

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TT

Trần Tuấn Hoàng

26 tháng 5 2022

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2n}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{n}\)

\(\Rightarrow2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{n}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2n}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2n}\)

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ND

Nguyễn Đoàn Hồng Thái

17 tháng 12 2017

a)Cho A= 1/2^2+1/3^2+...+1/n^2.CMR A<1

b)Cho B=1/2^2+1/4^2+1/6^2+...+1/(2n)^2.CMR B<1/2

c)Cho C=3/4+8/9+15/16+...+n^2-1/n^2.CMR C<n-2

#Toán lớp 6

0

NB

Nguyễn Bích Hằng

27 tháng 7 2017 - olm

RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

#Toán lớp 9

1

AN

alibaba nguyễn

27 tháng 7 2017

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

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T

TrầnHoàngGiang

25 tháng 8 2023 - olm

Bài 4: Tìm số tự nhiên n thoả mãn:

a) 4^n = 2^n+1

b) 16 = (n-1)^4

c) 125 = (2n+1)^3

#Toán lớp 6

2

ND

Nguyễn Đức Trí

VIP

25 tháng 8 2023

a) \(4^n=2^{n+1}\)

\(\Rightarrow2^{2n}=2^{n+1}\)

\(\Rightarrow2n=n+1\)

\(\Rightarrow n=1\)

b) \(16=\left(n-1\right)^4\)

\(\Rightarrow2^4=\left(n-1\right)^4\)

\(\Rightarrow n-1=2\)

\(\Rightarrow n=3\)

c) \(125=\left(2n+1\right)^3\)

\(\Rightarrow5^3=\left(2n+1\right)^3\)

\(\Rightarrow2n+1=5\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=2\)

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NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

25 tháng 8 2023

a, 4ⁿ = 2ⁿ⁺¹

(2²)ⁿ = 2ⁿ⁺¹

2²ⁿ = 2ⁿ⁺¹

2n = n + 1

2n - n = 1

n = 1

b, 16 = (n-1)⁴

2⁴ = (n-1)⁴

2 = n-1

n = 3

c, 125 = (2n + 1)³

5³ = (2n+1)³

5 = 2n + 1

2n = 4

n = 2

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TU

Toki Uni5

26 tháng 8 2018 - olm

Làm giúp nha

1.Tính nhanh các tổng sau

a) 1+2+3+4+....+n

b)1+3+5+7+.....+(2n-1)

c)2+4+6+8+.......+2n

d)1+6+11+16+.......+46+51

#Toán lớp 6

3

N

Nguyệt

26 tháng 8 2018

a)1+2+3+...+n

=[(n-1):1+1].(n+1):2

=n.( n+1)/2

b) {[(2n-1)-1]:2+1}. [(2n-1)+1]:2

=n.n=n²

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HT

Hacker_mũ trắng

17 tháng 1 2019

a) 1+2+3+...+n

= [(n-1):1+1].(n+1):2

= n.( n+1)/2

b) {[(2n-1)-1]:2+1}. [(2n-1)+1]:2

= n.n = n²

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NB

Nguyễn Bích Hằng

24 tháng 7 2017 - olm

1. Cho E= \(\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x+1\right)}}}{\sqrt{x^2-4\left(x+1\right)}}\) \(\left(1-\frac{1}{x-1}\right)\)
RG E.

2. Cho E= \(\frac{1+\sqrt{1-x}}{1-x+\sqrt{1-x}}+\frac{1-\sqrt{1+x}}{1+x-\sqrt{1+x}}+\frac{1}{\sqrt{1+x}}\)

a) RGBT
b) So sánh E với \(\frac{\sqrt{2}}{2}\)

#Toán lớp 9

0

MH

Minh Hoang Hai

6 tháng 7 2017 - olm

Rút gọn biểu thức :

a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{2^{2n}}\right)\)

b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)

#Toán lớp 8

0

MH

Minh Hoang Hai

6 tháng 7 2017

Rút gọn biểu thức :

a) \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{4}\right).\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)

b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)

#Toán lớp 8

1

NT

Nguyễn Thị Huyền Trang

6 tháng 7 2017

a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)

\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)

\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)

...

\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)

Vậy ...

b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)

Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)

\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)

\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)

\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)

...

\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)

Vậy ...

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NT

Nguyễn Thị Huyền Trang

6 tháng 7 2017

áp dụng hằng đẳng thức (a+b)(a-b)=a^2-b^2 Minh Hoang Hai

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NP

Nguyễn Phong

19 tháng 6 2019 - olm

Thu gọn biểu thức sau :

a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)

b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)

#Toán lớp 8

2

PT

Phạm Thị Thùy Linh

19 tháng 6 2019

\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=2^{64}-1-2^{64}=-1\)

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DL

Duc Loi

19 tháng 6 2019

a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)

Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)

\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)

b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.

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