2.

\(a,x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)

\(b,x^2-3y^2=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

\(c,\left(3x-2y\right)^2-\left(2x-3y\right)^2\\ =\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\\ =\left(x+y\right)\left(5x-5y\right)=5\left(x-y\right)\left(x+y\right)\)

\(d,9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)

\(e,\left(4x^2-4x+1\right)-\left(x+1\right)^2\\ =\left(2x-1\right)^2-\left(x+1\right)^2\\ =\left(2x-1-x-1\right)\left(2x-1+x+1\right)\\ =3x\left(x-2\right)\)

\(f,x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)

\(g,27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,027x+0,01\right)\)

\(h,125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\\ =\left(2-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}-\dfrac{2\left(3-\sqrt{3}\right)}{6}=\dfrac{12-6\sqrt{3}+3\sqrt{2}-6+2\sqrt{3}}{6}=\dfrac{6-\sqrt{3}}{6}\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\in Z\)

\(\Leftrightarrow2⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-4;-2;-1\right\}\\ \Leftrightarrow x\in\left\{1;4;16;25\right\}\)

Vậy \(x\in\left\{1;4;16;25\right\}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\)

Tick plz

\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

Tick plz

\(B=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{12}+\sqrt{20}}=\dfrac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}}{\sqrt{12}+\sqrt{20}}\\ B=\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{12}+\sqrt{20}}=\dfrac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{20}-\sqrt{12}\right)}{8}\\ B=\dfrac{2\sqrt{15}-6+10-2\sqrt{15}}{8}=\dfrac{4}{8}=\dfrac{1}{2}\)

Tick nha 😘

\(\sqrt{2x^2+4x+5}=\sqrt{2\left(x^2+2x+1\right)+3}=\sqrt{2\left(x+1\right)^2+3}\)

\(ĐKXĐ:2x^2+4x+5\ge0\Leftrightarrow2\left(x^2+2x+1\right)+3=2\left(x+1\right)^2+3\ge3>0\)(luôn đúng)

Vậy ĐKXĐ là \(x\in R\) hay pt luôn xác định với mọi x

\(\sqrt{x\left(x+2\right)}\)

\(ĐKXĐ:x\left(x+2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< -2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\ge0\\x< -2\end{matrix}\right.\)

\(a,P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\left(a\ne b;a,b\ge0\right)\)

\(P=\left(\dfrac{\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}-\sqrt{a}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)\cdot\dfrac{\sqrt{a}+\sqrt{b}}{b-\sqrt{ab}+a}\)

\(P=\left(\left(\sqrt{a}+\sqrt{b}\right)-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right)\cdot\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(P=\dfrac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

\(a,8+2\sqrt{15}=\left(\sqrt{3}+\sqrt{5}\right)^2\\ b,12+2\sqrt{35}=\left(\sqrt{7}+\sqrt{5}\right)^2\\ c,8+\sqrt{60}=8+2\sqrt{15}=\left(\sqrt{3}+\sqrt{5}\right)^2\)