\(2^{x+1}.3^y=12^x\)

\(2.2^x.3^y=2^{2x}.3^x\)

\(2.3^y=2^2.3^x\)

\(3^{y-x}=2\)

=> phương trình vô nghiệm

a) \(\dfrac{15}{11}-\left(\dfrac{5}{7}-\dfrac{18}{11}\right)+\dfrac{27}{7}=\dfrac{22}{7}+3=\dfrac{43}{77}\)

b) \(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{5}\right)=\dfrac{24}{5}+1=\dfrac{29}{5}\)

c) \(-1,2-0,8+0,25+5,75-2022=-2+6-2022=-2018\)

d) \(0,1+\dfrac{16}{9}+5,1+\dfrac{-20}{9}=\dfrac{26}{5}-\dfrac{4}{9}=\dfrac{214}{45}\)

\(A=3^{n+1}+9.3^{n+1}+2^n.4+2^n.8\)

\(=3^{n+1}.10+4.2^n.3\)

\(=3^n.6.5+2^n.2.6⋮6\)

\(\Rightarrow A⋮6\left(đpcm\right)\)

\(12+5\left(x-3\right)⋮\left(x-3\right)\) 

\(\Rightarrow12⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(12\right)=\left(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right)\)

Vì x>7 => x-3>4

\(\Rightarrow\left(x-3\right)\in\left\{6;12\right\}\)

\(\Rightarrow x\in\left\{9;15\right\}\)

Ta có:

\(\widehat{B}=180^o-90^o-30^o=60^o\)(tổng 3 góc trong tam giác)

\(AC=2BC\)(cạnh đối diện góc 30 độ)

Áp dụng định lý Pytago

\(AC^2=BC^2+AB^2\)

\(3BC^2=4\Rightarrow BC=\dfrac{2\sqrt{3}}{3}\)\(\Rightarrow AC=\dfrac{4\sqrt{3}}{3}\)

1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)

2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)

3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)

\(A=\dfrac{x^2-2x+7}{x^2-2x+3}=1+\dfrac{4}{x^2-2x+3}=1+\dfrac{4}{\left(x-1\right)^2+2}\)

\(A\in Z\Leftrightarrow\)\(\left[\left(x-1\right)^2+2\right]\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

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

\(\Leftrightarrow x=0\)

a) \(A=2\left(1+2+2^2+...+2^{59}\right)⋮2\)

b) \(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

c) \(A=2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^5+...+2^{58}\right)⋮7\)

\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)\(\left(đk:x\ne1;x\ge0\right)\)

\(=\dfrac{\sqrt{x}+1+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

Để P nguyên

=> \(\dfrac{1}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\left\{{}\begin{matrix}\text{x là số chính phương}\\\sqrt{x}+1\inƯ\left(1\right)=\left\{1;-1\right\}\end{matrix}\right.\)

\(\Rightarrow x=0\)

Xét tứ giác AMCD có:

ND=MN( giả thiết)

AN=NC(giả thiết)

=> tứ giác AMCD là hình bình hành