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}}

Bạn Detective_conan giải đúng đấy!

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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!

1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow A< 1\)

Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)

\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)

*Cách khác:

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Ta thấy:

\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)

\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Mà \(\frac{2+1}{2}< 3\)

\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)

tính bằng cách thuân tiện nhé

Ôi rồi ôi

câu này thì chịu 

Phương pháp giải:

Khi nhân hoặc chia một số với 1 thì giá trị số đó không thay đổi.

Lời giải chi tiết:

1 × 2 = 2          1 × 3 = 3  

2 × 1 = 2          3 × 1 = 3

2 : 1 = 2           3 : 1 = 3  

1 × 4 = 4          1 × 5 = 5

4 × 1 = 4          5 × 1 = 5

4 : 1 = 4           5 : 1 = 5

1 × 1 = 1          1 : 1 = 1

Ai gait hộ mình với . Mai mình phải nộp bài r.huhuhu

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) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)

a) 3 3/4 . x = 1 1/2

<=> 15/4 . x = 3/2

<=> x = 3/4 . 4/15

<=> x = 1/5

Vậy x = 1/5

b) 1 1/4 x + 1 1/2 = 1 1/4

<=> 5/4 . x + 3/2 = 5/4

<=> 5/4 . x = 5/4 - 3/2

<=> 5/4 . x = -1/4

<=> x = -1/4 . 4/5

<=> x = -1/5

Vậy x = -1/5

c) ( 3 1/3 - 1 1/2 x ) : 5/6 = 1 1/2

<=> ( 10/3 - 3/2 x ) : 5/6 = 3/2

<=> 10/3 - 3/2 x = 3/2 . 5/6

<=> 10/3 - 3/2 x = 5/4

<=> 3/2 . x = 10/3 - 5/4

<=> 3/2 . x = 25/12

<=> x = 25/12 . 2/3

<=> x = 25/18

Vậy x = 25/18

d) ( 3/7 x - 1 ) : 4 = -1/28

<=> 3/7 . x - 1 = -1/28 . 1/4

<=> 3/7 . x - 1 = -1/112

<=> 3/7 . x = -1/112 + 1

<=> 3/7 . x = 111/112

<=> x = 111/112 . 7/3

<=> x = 37/16

Vậy x = 37/16

e) | x - 3/4 | = 1

<=> x - 3/4 = 1

hoặc x - 3/4 = -1

<=> x = 1 + 3/4

hoặc x = -1 + 3/4

<=> x = 7/4

hoặc x = -1/4

Vậy x = 7/4 ; x = -1/4

f) | 2/3 . x + 1/3 | = 5/6

<=> 2/3 . x + 1/3 = 5/6

hoặc 2/3 . x + 1/3 = -5/6

<=> 2/3 . x = 5/6 - 1/3

hoặc 2/3 . x = -5/6 - 1/3

<=> 2/3 . x = 1/2

hoặc 2/3 . x = -7/6

<=> x = 1/2 . 3/2

hoặc x = -7/6 . 3/2

<=> x = 3/4

hoặc x = -7/4

Vậy x = 3/4 ; x = -7/4