Thực hiện phép trừ rồi điền kết quả vào chỗ trống.

4 - 1 = 3     4 - 2 = 2     3 + 1 = 4     1 + 2 = 3

3 - 1 = 2     3 - 2 = 1     4 - 3 = 1     3 - 1 = 2

2 - 1 = 1     4 - 3 = 1     4 - 1 = 3     3 - 2 = 1

dễ thế cũng hỏi

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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 \(\frac{2+1}{2}< 3\)

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

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

hơi sai đề

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!

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

Lời giải chi tiết:

4-1  >  2

4-2  =  2

4-3  <  4-2

4-1  <  3+1

Lời giải chi tiết:

Lời giải chi tiết:

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