\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{6}-\frac{1}{11}=\frac{5}{66}\)

= 9/10

k nha

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

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

\(=\frac{1}{10}\)

b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)

\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)

\(=\frac{1}{10}-\frac{1}{1000}\)

\(=\frac{99}{1000}\)

c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)

\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)

\(=4.\left(1-\frac{1}{90}\right)\)

\(=4.\frac{89}{90}\)

\(=\frac{178}{45}\)

_Chúc bạn học tốt_

a, \(=\frac{1}{10}\)

a) 1/10

b) ............

c)............

Mình giúp rồi t.i.c.k mình :v

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)

\(\Rightarrow5A=1-\frac{1}{8}\)

\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=5\left(1-\frac{1}{8}\right)\)

\(A=5.\frac{7}{8}\)

\(A=\frac{38}{8}\)

nhầm rồi :)

\(A=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}\right)\)

\(\Rightarrow A=\frac{5.7}{8}\)

\(\Rightarrow A=\frac{35}{8}\)

A  = 1 x 2 + 2 x 3 + ....... + 10 x 11

3A = 1 x 2 x 3 + 2 x 3 x 3 + ..........+ 10 x 11 x 3

3A = 1 x 2 x (3-0) + 2 x 3 x (4-1) + .......... + 10 x 11 x (12 -9)

3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ........... + 10 x 11 x 12 - 9 x 10 x 11

3A = (1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) +............ + 10 x 11 x 12

3A = 10 x 11 x 12 = 1320

A = 1320 : 3 = 440

Gọi biểu thức trên là A, ta có :

A= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300

 

= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12

= 1/3 - 1/12 = 1/4

Sai rồi bạn ạ

Phải là 1/3 nhân 4 chứ ko phải 1/3x4

a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)

\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)

\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)

b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)

\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)

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

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

\(=\frac{-9}{10}\)

sorry bn nha! mk ko bk lm phần c

trả lời :

a) 1/6

b)-9/10

Đặng Thị Dung cả 2 phần bạn CÔNG CHÚA ÔRI làm sai hết rồi. Nếu muốn biết sai ở đâu thì kb rồi nhắn tin mk chỉ cho

c,\(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}:\left(-1\frac{1}{4}\right)\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)

\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)

\(=\frac{1}{2}\times\left(-\frac{2}{3}\right)\times\frac{3}{4}\times...\times\left(-\frac{100}{101}\right)\)

=\(=\frac{1.\left(-2\right).3.....\left(-100\right)}{2.3.4.....101}=\frac{1}{101}\)

x la nhan ha ban

a)=1

b)2 phan 3