Ta có \(2^x-2^y=1024\Rightarrow x>y\)

Do đó \(2^y\left(2^{x-y}-1\right)=2^{10}\)

Lại có \(2^{x-y}-1\) lẻ và là ước 10 nên \(2^{x-y}-1=1\Rightarrow2^y=2^{10}\)

\(\Rightarrow y=10\Rightarrow2^{x-10}=2^1\Rightarrow x=11\)

Vậy \(\left(x;y\right)=\left(11;10\right)\)

\(a,\dfrac{1}{x}=\dfrac{1}{6}+3y\Leftrightarrow6=x+18xy\Leftrightarrow x\left(18y+1\right)=6\)

Mà \(x,y\in Z\)

Vậy ko có x,y nguyên tm

\(b,A=\dfrac{2\left(x+1\right)-3}{x+1}=2-\dfrac{3}{x+1}\in Z\\ \Leftrightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)

giải nghĩa tên ik mk làm cho :v

\(a,ĐK:x\ne-2;x\ne\pm\dfrac{1}{2}\\ b,B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\\ B=\dfrac{4\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)\left(x+2\right)}=\dfrac{4}{x+2}\\ c,\left|x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=4\\ \Leftrightarrow B=\dfrac{4}{4+2}=\dfrac{2}{3}\\ d,B\in Z\Leftrightarrow x+2\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-6;-4;-3;-1;0;2\right\}\)

\(\Leftrightarrow Mx^2=x^2-2x+\sqrt{2015}\\ \Leftrightarrow x^2\left(M-1\right)+2x-\sqrt{2015}=0\)

Ta có \(\Delta'\ge0\Leftrightarrow1+\sqrt{2015}\left(M-1\right)\ge0\)

\(\Leftrightarrow1+\sqrt{2015}M-\sqrt{2015}\ge0\\ \Leftrightarrow M\ge\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\)

Vậy \(M_{min}=\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\Leftrightarrow x=-\dfrac{b'}{a}=-\dfrac{1}{M-1}=\dfrac{-\sqrt{2015}}{\sqrt{2015}-1}\)

\(P=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(c,\) Vì AD//BP và AD=BP nên ADPB là hbh

Do đó O là trung điểm AP và BD

Xét tam giác ADP có DO và AN là trung tuyến giao tại G nên G là trọng tâm

Do đó \(DG=\dfrac{2}{3}DO\)

Mà \(DO=\dfrac{1}{2}BD\Rightarrow DG=\dfrac{2}{3}\cdot\dfrac{1}{2}BD=\dfrac{1}{3}BD\)