a) \(\frac{1}{9}\)

b) -1100

các bn ơi giải giúp mình đi mà

Câu 1;

\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)

Câu 2:

\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)

Câu 1;

13 −17 −113 23 −27 −213  ·34 −316 −364 −3256 1−14 −116 −164  +58 

=13 −17 −113 2(13 −17 −113 ) ·3(14 −116 −164 −1256 )4(14 −116 −164 −1256 ) +58 

=12 ·34 +58 =38 +58 =1

Câu 2:

rút gọn tu , bạn ơi

\(=1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+...+8+\frac{17}{2}\)

\(=\left(1+2+...+8\right)+\left(\frac{3}{2}+\frac{5}{2}+...+\frac{17}{2}\right)=36+\frac{80}{2}=36+40=76\)

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)

\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)

\(B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)

\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(B=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)

\(B=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)

Không thấy gì hết

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

\(A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+...+\frac{1+2+3+...+16}{16}\)

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

\(A=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{16+1}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)

\(A=\frac{2+3+4+5+...+17}{2}\)

\(A=\frac{152}{2}\)

\(A=76\)

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)

\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)

\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)

\(=\frac{3}{8}+\frac{5}{8}\)

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

\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8

=\(\frac{327}{722}\)+5/8

=\(\frac{3113}{2888}\)

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\) 

= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)

= \(\frac{15}{8}+\frac{5}{8}\)

= \(\frac{5}{2}\)

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)

\(=\frac{49}{34}\)

Xyz 

bạn sai từ dòng thứ 3 rồi

A=1+\(\frac{1+2}{2}\)+\(\frac{1+2+3}{3}\)+...+\(\frac{1+2+3+...+16}{16}\)

A=\(\frac{2}{2}\)+\(\frac{3}{2}\)+\(\frac{4}{2}\)+...+\(\frac{17}{2}\)

A=\(\frac{2+3+4+...+17}{2}\)

A=76(đề thi HSG huyện tui có tui làm zậy mà cũng có điểm tuyệt đối)

\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+3+....+16\right)\)

\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+\frac{1}{4}\cdot\frac{4.5}{2}+.....+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}\)

\(A=\frac{\left(2+3+4+.....+17\right)}{2}=\frac{\left(2+17\right).16}{2}=\frac{152}{2}=76\)

Ế, bạn Phạm Nguyễn Tất Đạt, bạn đâu có được điểm tuyệt đối

A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)

A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136

A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2

A = 1 + (3+4+5+...+17)/2

A = 1 + 150/2

A = 1 + 75

A = 76

A = 1 + 1/2.(1+2) + 1/3.(1+2+3) + 1/4.(1+2+3+4) + ...+ 1/16.(1+2+....+16)

A = 1 + 1/2.3 + 1/3.6 + 1/4.10 + ...+ 1/16.136

A = 1 + 3/2 + 4/2 + 5/2 + ....+ 17/2

A = 1 + (3+4+5+...+17)/2

A = 1 + 150/2

A = 1 + 75

A = 76

Ta có : \(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)...\frac{1}{16}.\left(1+2+3+...+16\right)\)

                \(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)

                  \(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

                  \(=\frac{2+3+4+...+17}{2}\)

                   \(=\frac{152}{2}\)

                    \(=76\)