Phần bể chứa nước sau hai lần chảy là:

\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\) (bể).

Phần bể chưa có nước là:

\(1-\dfrac{31}{35}=\dfrac{4}{35}\) (bể).

Đáp số: còn \(\dfrac{4}{35}\) bể chưa có nước.

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\)

\(2A=\frac{1}{3}-\frac{1}{57}\)

\(2A=\frac{6}{19}\)

\(A=\frac{3}{19}\)

Ta có:

A=1/15+1/35+1/63+1/99+...+1/2015+1/3135

   =1/3.5+1/5.7+1/7.9+1/9.11+...+1/45.47+1/47.49

   =1/3-1/5+1/5-1/7+1/7-1/9+...+1/45-1/47+1/47-1/49

   =1/3-1/49

   =49/147-3/147

   =47/147

\(A=\frac{1}{12}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2015}+\frac{1}{3135}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{45\cdot47}+\frac{1}{47\cdot49}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)

\(A=\frac{1}{3}-\frac{1}{49}\)

\(A=\frac{46}{147}\)

                             \(49-3=46\text{ nha bạn !}\)

\(3^{135}+2^{135}+3^{133}+2^{134}\)

\(=\left(3^{135}+3^{133}\right)+\left(2^{135}+2^{134}\right)\)

\(=3^{133}\cdot\left(3^2+1\right)+2^{133}\cdot\left(4+1\right)\)

\(=3^{133}\cdot10+2^{133}\cdot5\)

\(=5\cdot2\cdot\left(3^{133}+2^{132}\right)\)

\(=10\cdot\left(3^{133}+2^{132}\right)\)

\(3^{125}=3^{3.45}=\left(3^3\right)^{45}=27^{45}\)

\(5^{90}=5^{2.45}=\left(5^2\right)^{45}=25^{45}\)

\(27>25>0\Rightarrow27^{45}>25^{45}\)

\(\Rightarrow3^{135}>5^{90}\)