Lời giải:

Ta có:

\(\frac{1}{13}; \frac{1}{14}; \frac{1}{15}<\frac{1}{12}\)

\(\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}; \frac{1}{62};\frac{1}{63}< \frac{1}{60}\)

\(\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}\)

Do đó:

\(A< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{9}{20}+\frac{1}{20}\)

\(\Leftrightarrow A< \frac{1}{2}\) (đpcm)

Đặt biểu thức bằng A:

\(\Rightarrow A=\dfrac{1}{5}\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Ta thấy: \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{61}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{3}{31}+\dfrac{3}{61}< \dfrac{1}{2}\left(đpcm\right)\)

\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)

\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)

\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)

Ta có ; 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...

\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)

\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)

\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)

\(\Rightarrow B< \dfrac{2012}{2013}\)

Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)

\(\Rightarrow B< \dfrac{3}{4}\)

* Chắc vậy, sai thì thôg cảm ^^ * 

Còn j k hiểu thì ib nha

1, bạn xem lại đề 

2, 15(x-3) + 8x-21 = 12(x+1) +120 

<=> 23x - 66 = 12x + 132 

<=> 11x = 198 <=> x = 198/11 

3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4 

<=> 30x + 10 - 95 = 18x -12

<=> 12x = 73 <=> x = 73/12 

Gọi \(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{63}\) là \(S\)

\(S=1+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{63}+\dfrac{1}{64}\right)-\dfrac{1}{64}\\ =\left(1-\dfrac{1}{64}\right)+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{63}+\dfrac{1}{64}\right)\)

Ta nhận thấy:

\(\dfrac{1}{3}\) lớn hơn \(\dfrac{1}{4}\)

\(\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7}\) đều lớn hơn \(\dfrac{1}{8}\)

\(\dfrac{1}{9},\dfrac{1}{10},...,\dfrac{1}{15}\) đều lớn hơn \(\dfrac{1}{16}\)

\(\dfrac{1}{17},\dfrac{1}{18},...,\dfrac{1}{31}\) đều lớn hơn \(\dfrac{1}{32}\)

\(\dfrac{1}{33},\dfrac{1}{34},...,\dfrac{1}{63}\) đều lớn hơn \(\dfrac{1}{64}\)

\(\Rightarrow S>\left(1-\dfrac{1}{64}\right)+\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ S>\left(1-\dfrac{1}{64}\right)+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ S>\dfrac{63}{64}+\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)\\ S>\dfrac{63}{64}+3>3\)Mặt khác ta có:

\(S=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}\right)+\left(\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{63}\right)\)

\(\dfrac{1}{3}\) bé hơn \(\dfrac{1}{2}\)

\(\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7}\) đều bé hơn \(\dfrac{1}{4}\)

\(\dfrac{1}{9},\dfrac{1}{10},...,\dfrac{1}{15}\) đều bé hơn \(\dfrac{1}{8}\)

\(\dfrac{1}{17},\dfrac{1}{18},...,\dfrac{1}{31}\) đều bé hơn \(\dfrac{1}{16}\)

\(\dfrac{1}{33},\dfrac{1}{34},...,\dfrac{1}{63}\) đều bé hơn \(\dfrac{1}{32}\)

\(\Rightarrow S< 1+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)\\ S< 1+1+1+1+1+1\\ S< 6\)