\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

\(\Rightarrow\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< 1\left(đpcm\right)\)

ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

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

\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)

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

\(\Rightarrow A< 1\left(đpcm\right)\)

1/2+1/2 mũ 2+1/2 mũ 3+...+1/2 mũ 100

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

2A= 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\).

2A- A=( 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\))-(  \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+...+ \(\frac{1}{2^{99}}\)+ \(\frac{1}{2^{100}}\)).

A= 1- \(\frac{1}{2^{100}}\)< 1.

=> A< 1.

Vậy A< 1.

Ta có

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)

\(\Leftrightarrow2A=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+...+\frac{2}{2^{100}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

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

\(\Rightarrow A< 1\)

Vậy A<1 (đpcm)

Ta có:3.A=1+1/3+1/3^2+...+1/3^97 +1/3^98

=>3.A - A=(1+1/3+1/3^2+...+1/3^98 + 1/3^99)-(1/3+1/3^2 +1/3^3+...+1/3^98+1/3^99)

=>2.A=1-1/3^99

=>A=1/2 -1/3^99.1/2 <1/2

Vậy ... T I C K cho mình với nha

Câu 1:

b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)

\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)

c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)

Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)

\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)

Tinh 2A, roi lay 2A-A se chung to dc

Bạn xem lời giải của mình nhé:

Giải:

A luôn > 0 (vì các số hạng trong tổng A đều lớn hơn 0)(1)

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

\(A< 1\)(2)

Từ (1) và (2) \(\Rightarrow0< A< 1\left(đpcm\right)\)

Chúc bạn học tốt!

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