Question
(1 Điểm)
A 3135𝑏ể
B 3135
C 435𝑏ể
D 635𝑏ể
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\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\)
\(2A=\frac{1}{3}-\frac{1}{57}\)
\(2A=\frac{6}{19}\)
\(A=\frac{3}{19}\)
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
Ta có:
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
Coi \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\)
\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\right)\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}+\frac{2}{55.57}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(=\frac{1}{3}-\frac{1}{57}\)
\(=\frac{19}{57}-\frac{1}{57}=\frac{18}{570}=\frac{6}{19}\)
\(\Rightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
Vậy tổng trên bằng \(\frac{3}{19}\)
1/15 +1/35 +1/63 + 1/99 +...+1/2915 +1/3135
=1/3x5+1/5x7+1/7x9+....+1/53x55+1/55x57
=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/53-1/55+1/55-1/57
=1/3-1/57
=6/19 nhé
Ta có:1/15+1/35+1/63+1/99+...+1/2915+1/3135=1/3*5+1/5*7+1/7*9+1/9*11+...+1/53*55+1/55*57
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55+1/55-1/57
=1/3-1/57=19/57-1/57=18/57
\(3^{135}+2^{135}+3^{133}+2^{134}\)
\(=\left(3^{135}+3^{133}\right)+\left(2^{135}+2^{134}\right)\)
\(=3^{133}\cdot\left(3^2+1\right)+2^{133}\cdot\left(4+1\right)\)
\(=3^{133}\cdot10+2^{133}\cdot5\)
\(=5\cdot2\cdot\left(3^{133}+2^{132}\right)\)
\(=10\cdot\left(3^{133}+2^{132}\right)\)
\(3^{125}=3^{3.45}=\left(3^3\right)^{45}=27^{45}\)
\(5^{90}=5^{2.45}=\left(5^2\right)^{45}=25^{45}\)
\(27>25>0\Rightarrow27^{45}>25^{45}\)
\(\Rightarrow3^{135}>5^{90}\)
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.