A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\frac{8}{9}=0\)

8/9 -1/72-1/56-1/42-1/30-1/20-1/6-1/2 = \(\frac{1}{12}\)

k và kb nha ! ! ! animeboy

8/9 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2

= 8/9 - (1/2 + 1/6 + 1/12 + ... + 1/72)

= 8/9 - (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/8×9)

= 8/9 - (1 - 1/2 + 1/2 - 1/3 × 1/3 - 1/4 + ... + 1/8 - 1/9)

= 8/9 - (1 - 1/9)

= 8/9 - 8/9

= 0

^_^◆_◆☆_★^_-

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac 1{72}-\frac 1{56}-\frac 1 {30}-\frac 1{20}-\frac 1{12}-\frac 1{6}-\frac{1}2\)

\(=\frac{8}9-\frac{1}{8.9}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{8}9-\frac{1}9+\frac{1}{8}-....-\frac{1}2+1\)

\(=\frac 79+1= \frac{16}9\)

9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4

Giải :

ta có

  \(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)

=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)

=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)

=\(\frac{9}{10}-1+\frac{1}{10}=0\)                  (Mong online math ks cho mình nhé)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+.....+\frac{1}{2}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+........+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(\frac{9}{9}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)

= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)

= \(\frac{8}{9}-\frac{8}{9}\)

= \(0\)

"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).

0

Chúc bạn học tốt!!
 

2 2/9 - x = 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

20/9 - x = 1/3×4 + 1/4×5 + 1/5×6 + 1/6×7 + 1/7×8 + 1/8×9

20/9 - x = 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8 - 1/9

20/9 - x = 1/3 - 1/9

20/9 - x = 3/9 - 1/9

20/9 - x = 2/9

x = 20/9 - 2/9

x = 18/9 = 2

Vậy x = 2