37x8-9x14+100

= (36+1)x 8 - (8+1)x14 +25x4

= 36x8+8-8 x14- 14 +25x8

 =8x (36-14)+8-14+4x25

= 2x4x22+4x25- 6

= 4x(44+25-6

=4x69-6

=276-6=270

37x8-9x14+100 = (36+1)x 8 - (8+1)x14 +25x4 = 36x8+8-8 x14- 14 +25x8 =8x (36-14)+8-14+4x25 = 2x4x22+4x25- 6 = 4x(44+25-6 =4x69-6 =276-6=270

8x9+9x2+9x15-9x14-9

=9x(8+2+15-14-1)

=9x10

=90

8 x 9 + 9 x 2 + 9 x 15 - 9 x 14 - 9

=  8 x 9 + 9 x 2 + 9 x 15 - 9 x 14 - 9 x 1

=  9 x ( 8 + 2 + 15 - 14 -1 )

=  9 x         10

=  90

Đáp số: 90

\(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{54.59}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{54}-\frac{1}{59}=\frac{1}{4}-\frac{1}{59}=\frac{55}{236}\)

Đề bài?

Đúng jui đó bn Hạ Ninh !!!

Ý là đề bài cho biểu thức đó yêu cầu làm gì? Tìm x hay sao? Có cho điều kiện j hk?

\(A=\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+....+\frac{2}{81.85}\)

\(\Rightarrow2A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{81.85}\)

\(\Rightarrow2A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{85}\)

\(\Rightarrow2A=1-\frac{1}{85}\)

\(\Rightarrow A=\frac{84}{85}:2=\frac{42}{85}\)

tính A còn lại tự tính nha

a) A = 2/1x5 + 2/5x9 + 2/9x13 +....+2/81x85

\(\frac{2}{1x5}+\frac{2}{5x9}+\frac{2}{9x13}+...+\frac{2}{81x85}\)

\(A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{85}\)

\(A=1-\frac{1}{85}\)

\(\Rightarrow A=\frac{84}{85}\)

k nha

A = \(\frac{2}{1\cdot5}\) + \(\frac{2}{5\cdot9}\) + \(\frac{2}{9\cdot13}\) + ... + \(\frac{2}{81\cdot85}\)

   = \(\frac{1}{2}\) *  ( \(\frac{1}{1\cdot5}\) + \(\frac{1}{5\cdot9}\) + \(\frac{1}{9\cdot13}\) + ... + \(\frac{1}{81\cdot85}\) )

   = \(\frac{1}{2}\) * ( 1 - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{9}\) + \(\frac{1}{9}\) - \(\frac{1}{13}\) + ... + \(\frac{1}{81}\) - \(\frac{1}{85}\) )

   = \(\frac{1}{2}\) * ( 1 - \(\frac{1}{85}\) )

   = \(\frac{1}{2}\) * \(\frac{84}{85}\)

   = \(\frac{42}{85}\)

Đặt tổng trên là A

\(5A=\frac{5}{4x9}+\frac{5}{9x14}+\frac{5}{14x19}+...+\frac{5}{44x49}\)

\(5A=\frac{9-4}{4x9}+\frac{14-9}{9x14}+\frac{19-14}{14x19}+...+\frac{49-44}{44x49}\)

\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\)

\(5A=\frac{1}{4}-\frac{1}{49}\Rightarrow A=\frac{49-4}{4x5x49}=\frac{45}{4x5x49}=\frac{9}{4x49}\)