\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)

\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)

\(C=\frac{2}{3}\)

Bạn Cô nàng Thiên Bình làm đúng hết òi =.= 

a=7.[1/8+1/27-1/49]

   ------------------------

11.[1/8+1/27-1/49]

=7/11

cau b,c tuong tu nha h mk   

a)\(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)

\(A=\frac{7.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}\).

\(A=\frac{7}{11}\)

b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)

\(B=\frac{\frac{8}{9}.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}{4.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}\)

\(B=\frac{8}{9}:4=\frac{2}{9}\)

c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)\(C=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}\)

C=\(\frac{2}{3}\)

Mọi người đánh giúp mình nhé! Hạn là tối nay!!

\(\frac{3}{5}.\frac{8}{27}.\frac{5}{3}=1.\frac{8}{27}.1=\frac{8}{27}\)

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

\(\frac{12}{5}.4-4.\frac{7}{5}=4\left(\frac{12}{5}-\frac{7}{5}\right)=4.1=4\)

\(\frac{3}{5}x\frac{8}{27}x\frac{5}{3}\)

\(=\frac{3}{5}x\frac{5}{3}x\frac{8}{27}\)

\(=1x\frac{8}{27}\)

\(=\frac{8}{27}\)

\(\frac{7}{19}x\frac{1}{3}+\frac{7}{19}x\frac{2}{3}\)

\(=\frac{7}{19}x\left(\frac{1}{3}+\frac{2}{3}\right)\)

\(=\frac{7}{19}x1=\frac{7}{19}\)

\(\frac{12}{5}x4-4x\frac{7}{5}\)

\(=4x\left(\frac{12}{5}-\frac{7}{5}\right)\)

\(=4x1=4\)

Đúng luôn nên các bn nhớ k mk nhé

A) 3/5 x 8/27 x 5/3

= ( 3/5 x 8/27 ) x 5/3

= 8/45 x 5/3 

= 8/27

B) 7/19 x 1/3 + 7/19 x 2/3

= ( 7/19 x 1/3 ) + ( 7/19 x 2/3 )

= 7/57 + 14/57

= 7/19

C) 12/5 x 4 - 4 x 7/5

= ( 12/5 x 4 ) - ( 4 x 7/5 )

= 48/5 - 28/5

= 4

Hình như câu này tớ đã gặp đâu đó trong đề thi HSG rồi!

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)

\(=\frac{1}{2}\div4=\frac{1}{8}\)

câu này đơn giản lắm

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4+\frac{4}{7}+\frac{4}{49}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1}{2}:4=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\)

= \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\right):\frac{91}{80}\)

= \(\frac{1}{2}:4:\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

Gọi \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)

      \(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)

Từ đề bài ta có

\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)

\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)

\(D=182\times\frac{1}{8}\times\frac{80}{91}\)

\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)

cho tui nha

Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)

\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)

Vậy D=20
 

BẰNG 20 ĐÓ

\(\frac{12}{18}+\frac{1}{3}+\frac{1}{7}+\frac{2}{8}+\frac{27}{36}+\frac{42}{49}\)

\(=\frac{2}{3}+\frac{1}{3}+\frac{1}{7}+\frac{1}{4}+\frac{3}{4}+\frac{6}{7}\)

\(=\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)\)

\(=1+1+1\)

\(=3\)

Đặt P = ... ( biểu thức đề bài ) 

Nhận xét: Với \(k\inℕ^∗\) ta có: 

\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)

\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)