a: A={x∈N|x=3k+1; k∈N; 0<=k<=6}

b: B={x∈N|x=k³; 1<=k<=5}

a)Mỗi phần tử đều cách nhau 3 đơn vị

b)ko biết làm

??

2 bạn nhé.

\(\left(\dfrac{-4}{9}\right)^2=\dfrac{16}{81}\Rightarrow x=2\) 

\(\left(-\dfrac{1}{3}\right)^3=\dfrac{-1}{27}\Rightarrow x=1\)

\(\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\Rightarrow x=1\)

\(\left(-\dfrac{4}{9}\right)^x=\dfrac{16}{81}\\ \left(-\dfrac{4}{9}\right)^x=\left(\dfrac{4}{9}\right)^2\\ \left(-\dfrac{4}{9}\right)^x=\left(-\dfrac{4}{9}\right)^2\\ x=2\\ -----------\\ \left(-\dfrac{1}{3}\right)^{2x+1}=-\dfrac{1}{27}\\ \left(-\dfrac{1}{3}\right)^{2x+1}=\left(-\dfrac{1}{3}\right)^3\\ 2x+1=3\\ 2x=3-1=2\\ x=\dfrac{2}{2}=1\\ -----------\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\dfrac{1}{81}\\\left(-\dfrac{1}{3}\right)^{3x+1}=\left(\dfrac{1}{3}\right)^4\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\left(-\dfrac{1}{3}\right)^4\\ 3x+1=4\\ 3x=4-1=3\\ x=\dfrac{3}{3}=1\)

\(\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\Leftrightarrow\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\Leftrightarrow x=2\)

Không sử dụng dấu tương đương nhé!

c:

\(a,32< 2^n< 128\)

\(=>2^5< 2^n< 2^7\)

\(=>n=6\)

Vậy...

\(b,2.16\ge2^n>4\)

\(=>2^5\ge2^n>2^2\)

\(=>n\in\left\{3;4;5\right\}\)

Vậy...

\(c,3^2.3^n=3^5\)

        \(3^n=3^5:3^2\)

        \(3^n=3^3\)

\(=>n=3\)

Vậy...

\(d,\left(2^2:4\right).2^n=4\)

     \(\left(2^2:2^2\right).2^n=4\)

                 \(1.2^n=4\)

                    \(2^n=4:1\)

                    \(2^n=4\)

              \(=>2^n=2^2\)

             \(=>n=2\)

Vậy ...

\(e,\dfrac{1}{9}.3^4.3^n=3^7\)

   \(\dfrac{1}{9}.81.3^n=3^7\)

       \(3^2.3^n=3^7\)

           \(3^n=3^7:3^2\)

           \(3^n=3^5\)

\(=>n=5\)

Vậy...

\(g,\dfrac{1}{2}.2^n+4.2^n=9.2^5\)

 \(\left(\dfrac{1}{2}+4\right).2^n=9.2^5\)

             \(\dfrac{9}{2}.2^n=9.32\)

              \(\dfrac{9}{2}.2^n=288\)

                  \(2^n=288:\dfrac{9}{2}\)

                  \(2^n=2^6\)

\(=>n=6\)

Vậy...

a) \(32< 2^n< 128\\ \Rightarrow2^5< 2^n< 2^7\\ \Rightarrow5< n< 7\)

Mà: \(n\inℕ^∗\)

\(\Rightarrow n=6\)

b) \(2.16\ge2^n>4\\ \Rightarrow2^1.2^4\ge2^n>2^2\\ \Rightarrow2^5\ge2^n>2^2\\ \Rightarrow5\ge n>2\)

Mà: \(n\inℕ^∗\)

\(\Rightarrow n\in\left\{5;4;3\right\}\)

c) \(3^2.3^n=3^5\\ \Rightarrow3^{n+2}=3^5\\ \Rightarrow n+2=5\\ \Rightarrow n=3\left(nhận\right)\)

\(12=2^2.3\\ 20=2^2.5\)

\(\Rightarrow UCLN\left(12,20\right)=2^2=4\)

ƯCLN(12,20) = 4

Ta có : 1,  2, 3, 4, 5, 6, 7, 8, 9, 10...

           I,  II,  III, IV,  V,  VI, VII, VIII, IX, X...

Ta thấy: Để các chữ số gồm cả I và X không lặp lại quá hai lần thì có những số như:

IX, XI, XII, IXX ,XXI ,XXII 

Vậy ta có thể viết được: 6 số

\(#FallenAngel\)

IX,XI,XII,IXX,XXI,XXII

ĐKXĐ: \(x\notin\left\{1;-1;2;-2\right\}\)

\(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)

=>\(\dfrac{\left(x+4\right)\left(x+1\right)+\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+8\right)\left(x+2\right)+\left(x-8\right)\left(x+2\right)+6\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{x^2+10x+16+x^2-10x+16+6\left(x^2-4\right)}{x^2-4}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{2x^2+32+6x^2-24}{x^2-4}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{8x^2+8}{x^2-4}\)

=>\(\dfrac{x^2+4}{x^2-1}=\dfrac{4\left(x^2+1\right)}{x^2-4}\)

=>\(4\left(x^2+1\right)\left(x^2-1\right)=\left(x^2+4\right)\left(x^2-4\right)\)

=>\(4\left(x^4-1\right)=x^4-16\)

=>\(4x^4-4-x^4+16=0\)

=>\(3x^4+12=0\)(vô lý)

Vậy: Phương trình vô nghiệm

- Nếu x là số lẻ thì bó tay

- Nếu x là số chẵn: Đặt \(x=2k,n\inℕ\)

\(P=a^2a^4a^6...a^{2n}=a^{2+4+6+...+2n}=a^{42}\)

\(\Rightarrow2+4+6+...+2n=42\)

\(\Leftrightarrow2\left(1+2+3+...+n\right)=42\)

\(\Leftrightarrow\dfrac{2n\left(n+1\right)}{2}=42\)

\(\Leftrightarrow n\left(n+1\right)=42=6\times7\)

\(\Rightarrow n=6\Rightarrow x=12\)