\(a)\)\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(M=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{400-1}{400}\)

\(M=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{400}\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400}\right)\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Do từ 2 đến 20 có \(20-2+1=19\) nên : 

\(M=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\)

\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)

\(A>\frac{1}{2}-\frac{1}{21}\)

\(\Rightarrow\)\(M=19-A>19-\frac{1}{2}+\frac{1}{21}=18,5+\frac{1}{21}>8\)

\(\Rightarrow\)\(M>8\) ( đpcm ) 

Còn câu b) bn xem lại đề đi, nếu đề đúng thì mk sai :v 

Chúc bạn học tốt ~ 

\(M=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}...+\frac{399}{400}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+\left(1-\frac{1}{25}\right)+...+\left(1-\frac{1}{400}\right)\)

\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{20^2}\right)\)

\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2}\right)\)

Đặt \(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{20^2}\)

\(< P=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{20\cdot21}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}\)

\(\Rightarrow M+N>19-\frac{1}{2}+\frac{1}{21}=\frac{37}{2}+\frac{1}{21}>8\)

b sai  đề.chừng nào chữa đề thì làm

a: =6/12*7/14*8/16*9/18=1/2*1/2*1/2*1/2=1/16

b: =4/12*15/3*9/25*24/8

=1/3*3*5*9/25=9/5

A = 3/4 + 8/9 + 15/16 + ... + 399/400

A = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/400

A = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/400)

A = 19 - (1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20)

đặt b = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20

có 1/2.2 < 1/1.2 ; 1/3.3 < 1/2.3 ; ... 1/20.20 < 1/19.20

=> b < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/19.20

=> b < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/19 - 1/20

=> b < 1 - 1/20

=> b < 1

mà A = 19 - b

=> A > 18

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)

\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+.....+\frac{20^2-1}{20^2}\)

\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{20^2}\right)\)

\(>19-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{19\cdot20}\right)\)

\(=19-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{19}-\frac{1}{20}\right)\)

\(=19-\left(1-\frac{1}{20}\right)\)

\(>19-1=18\)

\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+...+\frac{20^2-1}{20^2}\)

\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+...+\frac{20^2}{20^2}-\frac{1}{20^2}\)

\(A=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)

Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(B< 1-\frac{1}{20}< 1\)

\(B< 1\)

\(\Rightarrow A>19-1\)

\(\Rightarrow A>18\)

=(1-1/2)(1+1/2)*...*(1+1/10)(1-1/10)

=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=1/10*11/2

=11/20

a,\(\frac{4}{3}\)x \(\frac{9}{8}\)x \(\frac{16}{15}\)x \(\frac{25}{24}\)

= \(\frac{5}{3}\)

b, \(\frac{4}{3}\)x \(\frac{9}{8}\)x \(\frac{16}{15}\)x \(\frac{25}{24}\)x \(\frac{36}{35}\)x \(\frac{49}{48}\)x \(\frac{64}{63}\)x \(\frac{81}{80}\)

= \(\frac{9}{5}\)

chi tiết hơn đc k bn

chi tiết hơn đi bạn

\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{10^2-1}{10^2}.\)

A là tổng của 9 số hạng; mỗi số hạng đều nhỏ hơn 1 nên A<9*1<50.

B. 18/24

b nhé