\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{399}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{7}{21}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\frac{6}{21}\)

\(=\frac{1}{7}\)

Chúc bạn học tốt !!! 

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{399}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

\(A=\frac{2}{7}:2=\frac{2}{7}.\frac{1}{2}=\frac{2}{14}=\frac{1}{7}\)

Trả lời

Đặt A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{399}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=\frac{1}{3}-\frac{1}{21}\)

\(2A=\frac{6}{21}\Rightarrow A=\frac{6}{21}:2\Rightarrow A=\frac{1}{7}\)

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\) 

A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)

A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)

A=\(\frac{1}{1}-\frac{1}{11}\)

=>A=\(\frac{10}{11}\)

B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\) 

2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

2B=\(\frac{1}{3}-\frac{1}{21}\)

2B=\(\frac{2}{7}\)

B=\(\frac{2}{7}:2\)

=>B=\(\frac{1}{7}\)

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)

\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(A=2.\frac{3}{16}\)

\(A=\frac{3}{8}\)

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)

\(2A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\)

\(2A=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)

\(2A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(2A=2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3}{8}\)

\(\Rightarrow A=\frac{3}{8}\div2=\frac{3}{16}\)

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{19.21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(\Rightarrow B=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

\(K=\left(1-\dfrac{3}{2\cdot4}\right)\left(1-\dfrac{3}{3\cdot5}\right)\cdot...\cdot\left(1-\dfrac{3}{19\cdot21}\right)\)

\(=\dfrac{3^2-1-3}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2-1-3}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{20^2-4}{\left(20-1\right)\left(20+1\right)}\)

\(=\dfrac{\left(3-2\right)\left(3+2\right)}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{\left(4-2\right)\left(4+2\right)}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{18\cdot22}{\left(20-1\right)\left(20+1\right)}\)

\(=\dfrac{1\cdot5}{2\cdot4}\cdot\dfrac{2\cdot6}{3\cdot5}\cdot...\cdot\dfrac{18\cdot22}{19\cdot21}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot21\cdot22}{2\cdot3\cdot4\cdot5\cdot...\cdot19\cdot20\cdot21}=1\cdot22=22\)