\(x\sqrt{x}-1=\sqrt{x}^3-1^3=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)-ĐK:x>=0\)

\(x+7\sqrt{x}-8=\left(x-\sqrt{x}\right)+\left(8\sqrt{x}-8\right)=\sqrt{x}\left(\sqrt{x}-1\right)+8\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+8\right)-ĐK:x>=0\)

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

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

\(\dfrac{9}{2\sqrt{3}}=\dfrac{9\sqrt{3}}{2\sqrt{3}^2}=\dfrac{9\sqrt{3}}{2.3}=\dfrac{3\sqrt{3}}{2}\)

\(\dfrac{5}{2-\sqrt{7}}=\dfrac{5\left(2+\sqrt{7}\right)}{\left(2-\sqrt{7}\right)\left(2+\sqrt{7}\right)}=\dfrac{5\left(2+\sqrt{7}\right)}{2^2-\sqrt{7}^2}=\dfrac{5\left(2+\sqrt{7}\right)}{4-7}=-\dfrac{5\left(2+\sqrt{7}\right)}{3}\)

\(\left(10x^3y^2-310x^2y^3+xy\right):2xy\\ =\dfrac{10x^3y^2}{2xy}-\dfrac{310x^2y^3}{2xy}+\dfrac{xy}{2xy}\\ =\dfrac{2xy.5x^2y}{2xy}-\dfrac{2xy.155xy^2}{2xy}+\dfrac{1}{2}\\ =5x^2y-155xy^2+\dfrac{1}{2}\)

\(ĐK:x\ne2\)

\(\)\(Để\ A\ là\ số\ nguyên\ dương\ thì:\)

\(\left\{{}\begin{matrix}\dfrac{4}{x-2}\in Z\left(1\right)\\\dfrac{4}{x-2}>0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)=>4⋮\left(x-2\right)=>x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\\ =>x\in\left\{3;1;4;0;6;-2\right\}\\ \left(2\right)=>x-2>0< =>x>2\)

\(=>\left\{{}\begin{matrix}x\in\left\{3;1;4;0;6;-2\right\}\\x>2\end{matrix}\right.< =>x\in\left\{3;4;6\right\}\left(TMDK\right)\)

\(2^{24}=2^{3.8}=\left(2^3\right)^8=\left(2.2.2\right)^8=8^8\\ 3^{16}=3^{2.8}=\left(3^2\right)^8=\left(3.3\right)^8=9^8\)

\(Vì :\) \(8^8< 9^8=>2^{24}< 3^{16}\)

\(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{5}\sqrt{9x-45}=3\left(ĐK:x>=5\right)\\ < =>\sqrt{x-5}+\sqrt{4}.\sqrt{x-5}-\dfrac{1}{5}.\sqrt{9}.\sqrt{x-5}=3\\ < =>\left(1+2-\dfrac{3}{5}\right).\sqrt{x-5}=3\\ < =>\dfrac{12}{5}.\sqrt{x-5}=3\\ < =>\sqrt{x-5}=\dfrac{5}{4}\\ < =>x-5=\dfrac{25}{16}\\ < =>x=\dfrac{105}{16}\left(TMDK\right)\)

\(\sqrt{25x^2-9}-2\sqrt{5x-3}=0\left(ĐK:x>=\dfrac{3}{5}\right)\\ < =>\sqrt{\left(5x-3\right)\left(5x+3\right)}-2\sqrt{5x-3}=0\\ < =>\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{5x-3}=0\\\sqrt{5x+3}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}5x-3=0\\5x+3=4\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{5}\left(TMDK\right)\\x=\dfrac{1}{5}\left(KTMDK\right)\end{matrix}\right.=>x=\dfrac{3}{5}\)

\(x+4\sqrt{x}=\sqrt{x}\left(\sqrt{x}+4\right)-ĐK:x>=0\)

\(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2-ĐK:x>=0\)

\(x\sqrt{x}+8=\sqrt{x}^3+2^3=\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)-ĐK:x>=0\)

\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\\ =\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=a-b=>A\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-10}{x-4}\left(x>=0;x\ne4\right)\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{x+2\sqrt{x}+x-3\sqrt{x}+2+\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{2x-8}{x-4}\\ =\dfrac{2\left(x-4\right)}{x-4}=2\)