\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)

\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\) 

Lời giải:

\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)

\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)

\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)

\(C=\dfrac{5}{5\cdot8\cdot11}+\dfrac{5}{8\cdot11\cdot14}+...+\dfrac{5}{302\cdot305\cdot308}\\ =\dfrac{5}{6}\cdot\left(\dfrac{6}{5\cdot8\cdot11}+\dfrac{6}{8\cdot11\cdot14}+...+\dfrac{6}{302\cdot305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{5\cdot8}-\dfrac{1}{8\cdot11}+\dfrac{1}{8\cdot11}-\dfrac{1}{11\cdot14}+...+\dfrac{1}{302\cdot305}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{40}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\dfrac{1}{40}-\dfrac{5}{6}\cdot\dfrac{1}{305\cdot308}\\ =\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}\\ \dfrac{5}{6\cdot305\cdot308}>0\Rightarrow\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}< \dfrac{1}{48}\)

a: \(x^2-6x+5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

b: \(x\left(x+5\right)+x\left(x+15\right)=0\)

\(\Leftrightarrow x\left(2x+20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)