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

\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)

\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^{101}}< \frac{1}{3}\)

\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)

\(\Rightarrow A< \frac{1}{2}\)

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

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

Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)

\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)

Vậy A<2

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

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

Ta đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Ta so sánh giữa A và C.

\(\frac{1}{3}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{3^3}< \frac{1}{3.4};....;\frac{1}{3^{99}}< \frac{1}{99.100}\Leftrightarrow A< C\)( 1 )

 \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

Mà \(\frac{99}{100}< \frac{1}{2}\Rightarrow C< B\)( 2 )

Từ ( 1 ) và ( 2 )

 \(\Rightarrow A< C< B\Leftrightarrow A< B\)

vế trước lớn hơn

1) \(+2x+3y⋮17\)

\(\Rightarrow26x+39y⋮17\)

\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)

Mà \(17x+34y⋮17\)

\(\Rightarrow9x+5y⋮17\)

\(+9x+5y⋮17\)

\(\Rightarrow36x+20y⋮17\)

\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)

Mà \(34x+17y⋮17\)

\(\Rightarrow2x+3y⋮17\)

Lời giải:

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{100}}$

$\Rightarrow 2A=1-\frac{1}{3^{100}}<1$

$\Rightarrow A< \frac{1}{2}$

$\Rightarrow A< B$