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

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

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

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

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

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

\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(=\frac{\frac{101.102}{2}}{51}\)

\(=101\)

\(S=1+2+2^2+...+2^{99}\)

\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)

\(S=3+2^2.3+...+2^{98}.3\)

\(=3\left(1+2^2+...+2^{98}\right)⋮3\)

dó là số 4

M=1784

3^x-1 + 5.3^x-1 = 162

 3^x-1. (1+5) = 162

3^x-1       =27

3^x-1        = 3^3

x-1          =3

x             = 4

Vậy x=4

\(3^{x-1}+5.3^{x-1}=162\)

\(\Rightarrow3^{x-1}.\left(1+5\right)=162\)

\(\Rightarrow3^{x-1}.6=162\)

\(\Rightarrow3^{x-1}=162:6=27=3^3\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x=3+1=4\)

chu so tc la3

cần câu trả lời cụ thể mà bạn

\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+1}\)

\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+1}=0\)

\(\left(x-2\right)^{x+1}\left(x-2-1\right)=0\)

\(\left(x-2\right)^{x+1}\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{x+1}=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy x = { 2; 3 }

\(\left(x-2\right)^{x+2}=\left(x-2\right)^{x+1}\)

\(\left(x-2\right)^{x+2}-\left(x-2\right)^{x+1}=0\)

\(\left(x-2\right)^{x+1}\left(x-2-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=2\\x=3\end{cases}}\)

a) Giả sử abcdeg chia hết cho 37                     —> 999abc+(abc+deg) chia hết cho 37

—> 999abc chia hết cho 37  vì 999 :37 ko dư                                                     —>abc + deg  chia hết cho 37