Câu 2:

\(a,\Leftrightarrow x=1;y=3\Leftrightarrow3=2+m-1\Leftrightarrow m=2\\ \text{Khi đó }y=2x+1\\ b,\text{PT hoành độ giao điểm: }2x+m-1=x-1\\ \Leftrightarrow m=-x\\ \text{Mà }x=0\Leftrightarrow m=0\)

Câu 4:

\(A=x+y+\dfrac{1}{2x}+\dfrac{2}{y}=\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)+\left(\dfrac{y}{2}+\dfrac{2}{y}\right)+\dfrac{1}{2}\left(x+y\right)\\ A\ge2\sqrt{\dfrac{x}{2}\cdot\dfrac{1}{2x}}+2\sqrt{\dfrac{y}{2}\cdot\dfrac{2}{y}}+\dfrac{1}{2}\cdot3=1+2+\dfrac{3}{2}=\dfrac{9}{2}\\ A_{min}=\dfrac{9}{2}\Leftrightarrow\left(x;y\right)=\left(1;2\right)\)

\(a,P=\dfrac{x^2-2x+2x+4-2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\\ b,\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-4\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{-4}{-4+2}=2\\ c,P=\dfrac{x+2-2}{x+2}=1-\dfrac{2}{x+2}\in Z\\ \Leftrightarrow x+2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow n\in\left\{-4;-3;-1;0\right\}\left(tm\right)\)

\(1,A=x^2+4x-5-x^2-4x=-5\\ B=x^2-2x+1-x^2+9=-2x+10\\ 2,\\ a,\Leftrightarrow x^2-x^2+x=15\Leftrightarrow x=15\\ b,\Leftrightarrow x^3-x^3-1=x\Leftrightarrow x=-1\)

\(ĐK:x>0;x\ne9\\ BPT\Leftrightarrow\dfrac{\sqrt{x}-1}{2}\cdot\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{2}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}-3\right)}-\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\\ \Leftrightarrow\dfrac{2}{2\left(\sqrt{x}-3\right)}< 0\Leftrightarrow\sqrt{x}-3< 0\left(2>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9\left(ĐKXĐ\right)\)

\(1,\\ A=\left(3-x\right)\left(3+x\right)\\ B=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-4+y\right)\\ 2,\\ a,\Leftrightarrow x\left(x^2-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\\ b,\Leftrightarrow x^2+3x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

\(N=3x-2y+5x-y-7y+2x=7x-10y\\ N=7\cdot2021-10\cdot2021=\left(-3\right)\cdot2021=-6063\)

\(=\sqrt{\left(\sqrt{17}-4\right)\left(\sqrt{17}+4\right)}=\sqrt{17-4^2}=1\)

Bài 1:

\(a,=\dfrac{4x-8}{x-2}=\dfrac{4\left(x-2\right)}{x-2}=4\\ c,=\dfrac{3x+1-2x}{x\left(x+1\right)}=\dfrac{x+1}{x\left(x+1\right)}=\dfrac{1}{x}\\ e,=\dfrac{2x-3+2x^2+2x-3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+x}{\left(x-1\right)\left(x+1\right)}\\ b,=\dfrac{x-13+x+3}{x-5}=\dfrac{2\left(x-5\right)}{x-5}=2\\ d,=\dfrac{x+4+x-4-3x}{x\left(x-4\right)}=\dfrac{-x}{x\left(x-4\right)}=\dfrac{1}{4-x}\\ f,=\dfrac{x+8-16+2x-16}{\left(x-8\right)\left(x+8\right)}=\dfrac{3x-24}{\left(x-8\right)\left(x+8\right)}=\dfrac{3\left(x-8\right)}{\left(x-8\right)\left(x+8\right)}=\dfrac{3}{x+8}\)

\(Q=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2\left(-6\right)=13\\ P=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ P=\left(-1\right)^3-3\left(-6\right)\left(-1\right)=-1-18=-19\)

\(=\left|\sqrt{5}-2\right|-\dfrac{\sqrt{5}+2}{5-4}=\sqrt{5}-2-\sqrt{5}-2=-4\)