Mọi ng giúp mình nha mình K cho!

khó thế?

Khó thế nhỉ 

\(\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)

=\(3.\left(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)

=\(\frac{3}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\right)\)

=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)

=\(\frac{3}{2}.\left(\frac{7}{28}-\frac{2}{28}\right)\)

=\(\frac{3}{2}.\frac{5}{28}=\frac{15}{56}\)

\(\sqrt[]{\frac{ }{ }\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^2}\)

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

a) Ta có:

\(\frac{11a+3b}{11c+3d}=\frac{11bk+3b}{11dk+3d}=\frac{b\left(11k+3\right)}{d\left(11k+3\right)}=\frac{b}{d}\) (1)

\(\frac{3a-11b}{3c-11d}=\frac{3bk-11b}{3dk-11d}=\frac{b\left(3k-11\right)}{d\left(3k-11\right)}=\frac{b}{d}\) (2)

Từ (1) và (2) suy ra \(\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}\) (đpcm)

b) Ta có:

\(\frac{1111c-99d}{9999c-11d}=\frac{1111dk-99d}{9999dk-11d}=\frac{d\left(1111k-99\right)}{d\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\) (1)

\(\frac{1111a-99b}{9999a-11b}=\frac{1111bk-99b}{9999bk-11b}=\frac{b\left(1111k-99\right)}{b\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\) (2)

Từ (1) và (2) suy ra \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\) (đpcm)

\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)

\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)

\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)

giup toi

\(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)

\(\frac{3}{4}x-\frac{44}{3}:\left(\frac{11}{15}+\frac{11}{35}+\frac{11}{63}\right)=12\)

\(\frac{3}{4}x-\frac{44}{3}:\frac{11}{9}=12\)

\(\frac{3}{4}x-12=12\)

\(\frac{3}{4}x=12+12\)

\(\frac{3}{4}x=24\)

\(x=24:\frac{3}{4}\)

\(x=32\)

vậy \(x=32\)

ket qua bang 32

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

\(2S=\frac{1}{2}-\frac{1}{10}\)

\(2S=\frac{2}{5}\)

\(S=\frac{2}{5}:2\)

\(S=\frac{1}{5}\)

S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(\Rightarrow2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{2}{5}:2=\frac{2}{5}.\frac{1}{2}=\frac{1}{5}\)