a) Ta có :

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

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

\(\Rightarrow\)A < 1 

b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(B=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{n^2}\right)\)

vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)

\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)

cảm ơn nha!

\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^4=17\)

\(\left(\frac{1}{2}\right)^x.\left[1+\left(\frac{1}{2}\right)^4\right]=17\)

\(\left(\frac{1}{2}\right)^x.\frac{17}{16}=17\)

\(\left(\frac{1}{2}\right)^x=17:\frac{17}{16}\)

\(\left(\frac{1}{2}\right)^x=16\)

\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^{-4}\)

\(\Rightarrow\)x = -4

Vậy x = -4

Số cần tìm là :

      123 x 123 x 123 = 1860867

               Đ/s:.....

CBHT!!! tk mk nha

~ thank you so much ~

Số cần tìm là

  123 x 123 x 23 = 347967

    đáp số 347967