Ta suy ra được:

\(\frac{3}{2}+\frac{3}{14}+\frac{3}{35}+\frac{3}{65}+...+\frac{3}{\frac{(x-3)x}{2}}=\frac{96}{49}\)

Vậy xét về mặt tổng quát ta sẽ quy đổi về công thức \(\frac{3}{\frac{(x-3)x}{2}}\).

Ta có:

\(\frac{3}{2}=\frac{3}{\frac{4}{2}}=\frac{3}{\frac{4.1}{2}}=\frac{4-1}{\frac{4.(4-3)}{2}}\)

Tương tự với các số còn lại. Sau đó ta có được kết quả sau:

\(\big(2-\frac{1}{2}\big)+\big(\frac{1}{2}-\frac{2}{7}\big)+\big(\frac{2}{7}-\frac{1}{5}\big)+\big(\frac{1}{5}-\frac{2}{13}\big)+\cdots+\big(\frac{2}{x-3}-\frac{2}{x}\big)=\frac{96}{49}\)

\(\Leftrightarrow2-\frac{2}{x}=\frac{96}{49}\)

\(\Leftrightarrow x=49\)

a,

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)

b,

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)

c, (Chịu :V)

d,

\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)

Chúc bạn học tốt nha.

1000010000803

chúc bạn học tốt~

Sai rồi họk toán vs đamê à

dãy số dài như thế này dễ lẫn lắm bạn ạ

a)\(27.65+27.35+300=27.\left(65+35\right)+300\)

\(=27.100+300=2700+300=3000\)

b)\(3838:\left[\left(190-6.5^2\right):4+3\right]\)

\(=3838:\left[\left(190-6.25\right):4+3\right]\)

\(=3838:\left[\left(190-150\right):4+3\right]\)

\(=3838:\left[40:4+3\right]=3838:\left[10+3\right]\)

\(=3838:13=\dfrac{3838}{13}\)

c)\(2022-x=2021\)

\(x=2022-2021=1\)

d)\(26+14:\left(x-5\right)=33\)

\(14:\left(x-5\right)=33-26=7\)

\(x-5=14+7=2\)

\(x=2+5=7\)

e)đề hỏi làm j thế bạn

a) \(227+50+23=\left(227+23\right)+50=250+50=300\)

b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)

c) \(1+2+3+4+5+...+97+98+99+100\)

\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)

\(=101+101+101+...+101\)

\(=101\cdot50\)

\(\Leftrightarrow5050\)

d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)

e) \(50-49+48-47+...+4-3+2-1\)

\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)

\(=1+1+1+1+..+1\)

\(=1\cdot25\)

\(=25\)

f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)

\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)

\(=10000\cdot360\)

\(=3600000\)

g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)

h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)

i) \(13\cdot17-256:16+14:7-2021^0\)

\(=13\cdot17-4^4:4^2+2-1\)

\(=13\cdot17-16+2-1\)

\(=13\cdot17-17\)

\(=17\cdot\left(13-1\right)\)

\(=204\)

j) \(7^2-36:3=49-12=37\)

 \(\frac{3}{2}+\frac{3}{14}+\frac{3}{15}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

 \(\frac{6}{\left(1.2\right).2}+\frac{6}{\left(2.7\right).2}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

 \(\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(x-3\right).x}=\frac{96}{49.2}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{96}{98}\)

=> \(1-\frac{1}{x}=\frac{48}{49}\)

=> \(\frac{1}{x}=\frac{1}{49}\)

=> \(x=49\)

nhớ giải bằng cách hợp lí ra nhé

Viết rối qá chả thấy j.

\(99^2vs9999^{10}\)

\(9999^{10}=\left(101\cdot99\right)^{10}=101^{10}\cdot99^{10}\)

Vì \(99^{10}>99^2=>99^2< 9999^{10}\)

a) Ta có: 2^91 = (2^13)^7 = 8192^7

5^35 = (5^5)^7 = 3125^7

Vì 8192 > 3125 nên 8192^7 > 3125^7

Vậy 2^91 > 5^35

b) Ta có: 9999^10 = 99^10 . 101^10

Vì 99^2 < 99^10 nên 99^2 < 99^10 . 101^10

Vậy 99^2 < 9999^10

c) Ta có: 2^300 = (2^6)^50 = 64^50

3^200 = (3^4)^50 = 81^50

Vì 49 < 64 < 81 nên 49^50 < 64^50 < 81^50

Vậy 49^50 < 2^300 < 3^200

d) 9^3/25^3 = (9/25)^3

3^6/2^12 = (3^2)^3/(2^4)^3 = 9^3/16^3 = (9/16)^3

Vì 9/25 < 9/16 nên (9/25)^3 < (9/16)^3

Vậy 9^3/25^3 < 3^6/2^12.