Ta có:

\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}+\dfrac{1}{8.10}\)

\(=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\right)\) \(+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\right)\)

Đặt \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{7.9}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)=\dfrac{4}{9}\)

Đặt \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{8.10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{5}\)

\(\Rightarrow S=A+B=\dfrac{4}{9}+\dfrac{1}{5}=\dfrac{29}{45}\)

Vậy \(S=\dfrac{29}{45}\)

Ta có:

\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}.\dfrac{50}{51}=\dfrac{25}{17}\)

Giải:

a) \(\Delta ALC\) vuông tại \(L\) ta có:

\(\cos A=\dfrac{AL}{AC}\left(1\right)\)

\(\Delta ANB\) vuông tại \(N\) ta có:

\(\cos A=\dfrac{AN}{AB}\left(2\right)\) Hay \(AN=AB.\cos A\left(3\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{AL}{AC}=\dfrac{AN}{AB}\\\text{A: chung}\end{matrix}\right.\)

\(\Rightarrow\Delta ANL\) đồng dạng với \(\Delta ABC\left(c-g-c\right)\) (Đpcm)

b) \(\Delta BLC\) vuông tại \(L\) ta có:

\(BL=BC.\cos B\left(4\right)\)

\(\Delta AMC\) vuông tại \(M\) ta có:

\(CM=AC.\cos C\left(5\right)\)

Từ \(\left(3\right);\left(4\right)\) và \(\left(5\right)\) suy ra:

\(AN.BL.CM=AB.\cos A.BC.\cos B.CA.\cos C\)

Hay \(AN.BL.CM=AB.BC.CA.\cos A.\cos B.\cos C\) (Đpcm)

x = 1

b) x = 7/3 hoặc x = 8/3                  

Hợp số: 4;6;8;9

Số nguyên tố : 3;5;7
 

Giải:

Theo đề bài: \(8b-9a=31\)

\(\Rightarrow b=\dfrac{31+9a}{8}=\dfrac{32-1+8a+a}{8}\)

\(=\left[\left(4+a\right)+\dfrac{a-1}{8}\right]\in N\Leftrightarrow\dfrac{a-1}{8}\in N\)

\(\Leftrightarrow\left(a-1\right)⋮8\Leftrightarrow a=8k+1\left(k\in N\right)\)

Khi đó:

\(b=\dfrac{31+9\left(8k+1\right)}{8}=9k+5\) \(\Rightarrow\dfrac{11}{17}< \dfrac{8k+1}{9k+5}< \dfrac{23}{29}\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left(9k+5\right)< 17\left(8k+1\right)\Leftrightarrow k>1\\29\left(8k+1\right)< 23\left(9k+5\right)\Leftrightarrow k< 4\end{matrix}\right.\)

\(\Leftrightarrow k\in\left\{2;3\right\}\)

Với \(\left[{}\begin{matrix}k=2\Rightarrow\left\{{}\begin{matrix}a=17\\b=23\end{matrix}\right.\\k=3\Rightarrow\left\{{}\begin{matrix}a=25\\b=32\end{matrix}\right.\end{matrix}\right.\)

Vậy...