\(2^2.3^{2n}.\left(\frac{2}{3}\right)^n.2^n=82944\)

\(2^2.9^n.\left(\frac{2}{3}\right)^n.2^n=2^{10}.3^4\)

\(2^2.2^n.\left(\frac{2}{3}.9\right)^n=2^{10}.3^4\)

\(2^{n+2}.6^n=2^{10}.3^4\)

\(2^{n+2}.2^n.3^n=2^{10}.3^4\)

\(2^{2n+2}.3^n=2^{10}.3^4\)

Vậy n = 4

a) S hình thoi là:

      (19 x 12) : 2 = 114(cm2)

b) S hình thoi là;

      (30 x 7) : 2 = 105(cm2)

\(2^n.3^{2n}.\left(\frac{2}{3}\right)^n.2^n=82944\)(n\(\in\)N)

\(2^n.2^n.\left(\frac{2}{3}\right)^n.\left(3^2\right)^n=82944\)

\(\left(2.2.\frac{2}{3}.9\right)^n=82944\)

\(24^n=82944\)

Tớ làm đến đây thôi khó lắm bạn xem lại đề đi

sorry, em mới học lớp 6 thi à

Mỗi lần bạn lên OLM là toàn đang những câu hỏi cực khó

Lời giải:

Xét số hạng tổng quát:

\(\frac{2n+1}{[n(n+1)]^2}=\frac{1}{n(n+1)}.\frac{2n+1}{n(n+1)}=\frac{n+1-n}{n(n+1)}.\frac{n+(n+1)}{n(n+1)}\)

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

Do đó:

\(S=\frac{3}{(1.2)^2}+\frac{5}{(2.3)^2}+....+\frac{2n+1}{[n(n+1)]^2}\)

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

\(=1-\frac{1}{(n+1)^2}\)

Ta có:

\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}=\frac{\left(1.3.5...2n-1\right).\left(2.4.6...2n\right)}{\left(2.4.6...2n\right)\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)

                                \(=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{1.2.3...n\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n.2^n}\)

                                \(=\frac{1}{2^n}\)

help meeeeeeeeeee