kinh q u e

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Dấu chấm là nhân

a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)

CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!

Tính nhanh mỗi biểu thức sau:

a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10

= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10

= 20 x 10 + 10

= 200 + 10

= 210

b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0

= A x 0

= 0

c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 0 : A

= 0

d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= (30 + 7 - 37)  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= (37 - 37)  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= 0  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= 0 x A

= 0

e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)

= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)

= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)

= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)

= 0 : (1 x 2 x 3 x 4 x ... x 10)

= 0 : A

= 0

g, (m : 1 - m x 1) : (m x 2008 + m x 2008)

= (m - m) : (m x 2008 + m x 2008)

= 0 : (m x 2008 + m x 2008)

= 0 : A

= 0

h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)

= (2 + 4 + 6 + 8 + m x n) x (972 - 972)

= (2 + 4 + 6 + 8 + m x n) x 0

= A x 0

= 0

l, (1 + 2 + 3 + ... + 99)  x (13 x 15 - 12 x 15 - 15)

= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))

= (1 + 2 + 3 + ... + 99) x (15 x 0)

= (1 + 2 + 3 + ... + 99) x 0

= A x 0

= 0

i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)

= 0 x : (2 + 4 + 6 +...+ 98)

= 0 x A

= 0

k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)

= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))

= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)

= (0 + 1 + 2 +...+ 97 + 99) x 0

= A x 0

= 0

GIÚP MÌNH VỚI  

(-7) NHÂN (-24) + (-36) : (-3)^2 - (-5)^3

1+2+3+4+5+6+7+8+9+...........+99=X+1+2+3+4+5+6+7+8+9+.................+99

4950=4950+X

X=4950-4950

X=0

bằng 0 nha bn

các bạn giúp mình với a. Mình cảm ơn trước

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)

Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)

\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

Vậy x = -100

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{99}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{16}{99}\)

\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}=\frac{32}{99}\)

=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)

=> \(\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)

=> \(\frac{1}{x+2}=\frac{1}{99}\)

=> x + 2 = 99

=> x = 97

Vậy x = 97 là giá trị cần tìm 

\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{x\times\left(x+2\right)}\)

\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{x\times\left(x+2\right)}\right)\)

\(=\frac{1}{2}\times\left(\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+...+\frac{x+2-x}{x\times\left(x+2\right)}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{x+2}\right)\)

\(=\frac{1}{6}-\frac{1}{2\times\left(x+2\right)}=\frac{16}{99}\)

\(\Leftrightarrow\frac{1}{2\times\left(x+2\right)}=\frac{1}{6}-\frac{16}{99}=\frac{1}{198}\)

\(\Leftrightarrow2\times\left(x+2\right)=198\)

\(\Leftrightarrow x+2=99\)

\(\Leftrightarrow x=97\)