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

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

Ta thấy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}<1\)

=>A>19-1=18(đpcm)

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

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

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

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

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{20^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{19\cdot20}\)

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

\(=\frac{19}{20}< 1\)

\(\Rightarrow A< 1\)

\(\Rightarrow M>18\)

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

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

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

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

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

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

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

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

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

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

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

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

.......

\(\frac{1}{20^2}< \frac{1}{19.20}=\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(\Rightarrow A>20-1=19\) (2)

Từ (1) và (2) => 19 < A < 20 

Vậy...

số số hạng là 19 chứ ko phải 20 ST

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

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

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

\(B< 1\)

\(\Rightarrow A>19-1\)

\(\Rightarrow A>18\)

\(=1+\frac{1}{3}+1+\frac{1}{15}+...+1+\frac{1}{399}.\)

\(=10+\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

=\(10+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)

=\(10+\frac{1}{2}\left(1-\frac{1}{21}\right)=10+\frac{1}{2}.\frac{20}{21}=\frac{220}{21}\)

Thanh you

Tham khảo tại đây:

Câu hỏi của triệu minh Anh - Toán lớp 6 - Học toán với OnlineMath

ko ngờ đấy mày lại ko được giải khi thi MYTS
 

MYTS  là j ạ

ko jup thì thôi đăng làm j