câu g) 

\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)

\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)

\(=\frac{12}{3}=4\)

câu mình trả lời sai rồi thông cảm

\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)

\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)

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

\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)

\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)

Bài 1: a, Tìm số nguyên a để tích hai phân số:

-19/5 và a/a-1 là một số nguyên

Tích của -19/5 và a/a-1 là:

-19/5 x a/a-1 = \(\frac{-19a}{5\left(a-1\right)}\)

A = \(\frac{-19a}{5\left(a-1\right)}\) là số nguyên khi và chỉ khi:

-19a ⋮ [5.(a - 1)]

[5.19.a - 5.19 + 5.19] ⋮ [(5.(a-1)]

[19.5(a - 1) + 5.19] ⋮ [5.(a - 1)]

5.19 ⋮ 5.(a - 1)

(a - 1) ∈ Ư(19) = {-19; -1; 1; 19}

a ∈ {-18; 0; 2; 20}

a = - 18 thì A = \(\frac{-19}{5}\times\frac{-18}{-18-1}\) = - 18/5 (loại)

a = 0 thì A =m -19/5 x 0/0-1 = 0 (nhận)

a = 2 thì A = - 19/5 x 1/2-1 = -38/5 (loại)

a = 20 thì A = -19/5 x 20/(20 - 1) = - 19/5 - 4

Vậy a ∈ {0; 20}

Bài đầu đơn giản rồi , tự tính nhé <3

Bài 2

\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n.3^2-2^n.2^2+3^n-2^n\)

\(=\left(3^n.3^2+1\right)-\left(2^n.2^2+1\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10.\left(3^n-2^{n-1}\right)⋮10\)

Vậy.....

a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)

\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)

\(\Leftrightarrow3n-1=2\)

\(\Leftrightarrow3n=3\)

\(\Leftrightarrow n=1\)

b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)

\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)

\(\Leftrightarrow n+2=-1\)

\(\Leftrightarrow n=-3\)

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

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

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

\(\Leftrightarrow-n+1=-3\)

\(\Leftrightarrow n=-4\)

c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)

\(\Leftrightarrow3n+1=-3\)

\(\Leftrightarrow3n=-4\)

\(\Leftrightarrow n=-\frac{4}{3}\)

\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)

    \(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)

     \(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)

      \(=\frac{1}{2003}\)

\(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\)

\(x+\frac{1}{8}=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{1}{8}\)

\(x=\frac{4}{16}-\frac{2}{16}\)

\(x=\frac{1}{8}\)

Vậy \(x=\frac{1}{8}\)

b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\)

      \(\frac{8}{27}-x=\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{9}{27}\)

                     \(x=-\frac{1}{27}\)

Vậy \(x=-\frac{1}{27}\)

c) \(x.\left(-\frac{1}{2}\right)^4=\frac{3}{8}\)

 \(x.\frac{1}{16}=\frac{3}{8}\)

       \(x=\frac{3}{8}:\frac{1}{16}\)

        \(x=\frac{3}{8}.16\)

      \(x=6\)

c) \(\left(\frac{1}{2}\right)^3.x=\left(\frac{1}{2}\right)^5\)

\(x=\left(\frac{1}{2}\right)^5:\left(\frac{1}{2}\right)^3\)

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

\(x=\frac{1}{4}\)

Vậy \(x=\frac{1}{4}\)

Chúc bạn học tốt !!!

a) \(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\Leftrightarrow x+\frac{1}{8}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}-\frac{1}{8}\Leftrightarrow x=\frac{1}{8}\)

b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\Leftrightarrow\frac{8}{27}-x=\frac{1}{3}\Leftrightarrow-x=\frac{1}{3}-\frac{8}{27}\Leftrightarrow-x=\frac{1}{27}\Leftrightarrow x=-\frac{1}{27}\)

c) \(x.\left(\frac{-1}{2}\right)^4=\frac{3}{8}\Leftrightarrow x.\frac{1}{16}=\frac{3}{8}\Leftrightarrow x=\frac{3}{8}:\frac{1}{16}\Leftrightarrow x=6\)

d) \(\left(\frac{1}{2}\right)^2.x=\left(\frac{1}{2}\right)^5\Leftrightarrow\frac{1}{8}.x=\frac{1}{32}\Leftrightarrow x=\frac{1}{32}:\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)