\(\left(e\right):\left(x^2+2xy+y^2\right)-3x-3y\\ =\left(x+y\right)^2-3\left(x+y\right)=\left(x+y\right)\left(x+y-3\right)\\ \left(g\right):xy\left(x+y\right)+4x+4y=xy\left(x+y\right)+4\left(x+y\right)\\ =\left(x+y\right)\left(xy+4\right)\\ \left(h\right):x^3-x+3x^2y+3xy^2+y^3-y=\left(x+y\right)^3-\left(x+y\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)

\(-\dfrac{5}{11}+\left(-\dfrac{6}{11}+1\right)=-\dfrac{5}{11}-\dfrac{6}{11}+1\\ =\dfrac{-5-6}{11}+1=-\dfrac{11}{11}+1\\ =-1+1=0\)

\(\left(a\right):3x-12xy=3x.1-3x.4y=3x.\left(1-4y\right)\\ \left(b\right):9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\\ \left(c\right):4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\\ \left(d\right):\left(3x+1\right)^2-\left(x+1\right)^2=\left[\left(3x+1\right)+\left(x+1\right)\right].\left[\left(3x+1\right)-\left(x+1\right)\right]=\left(4x+2\right).2x=4x\left(2x+1\right)\)

\(\left(2x-y\right)^3-8-3\left(x-y\right)\left(2x-y-2\right)\\ =\left(2x-y-2\right)\left[\left(2x-y\right)^2+2\left(2x-y\right)+2^2\right]-\left(3x-3y\right)\left(2x-y-2\right)\\ =\left(2x-y-2\right)\left(4x^2-4xy+y^2+4x-2y+4-3x+3y\right)\\ =\left(2x-y-2\right)\left(4x^2-4xy+y^2+x+y+4\right)\)

\(VP=a^3+b^3+c^3-3abc\\ =\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc\\ =\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VT\left(DPCM\right)\)

\(\left(f\right):x^3-5x^2+x-5=0\\ < =>x^2\left(x-5\right)+\left(x-5\right)=0\\ < =>\left(x-5\right)\left(x^2+1\right)=0\\ =>\left[{}\begin{matrix}x-5=0\\x^2+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x^2=-1< 0\left(KTM\right)\end{matrix}\right.=>S=\left\{5\right\}\\ \left(h\right):x\left(x^2+1\right)-2x\left(x-2\right)-2\left(x^2+1\right)=0\\ < =>\left(x^2+1\right)\left(x-2\right)-2x\left(x-2\right)=0\\ < =>\left(x-2\right)\left(x^2+1-2x\right)=0\\ =>\left[{}\begin{matrix}x-2=0\\\left(x-1\right)^2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.=>S=\left\{2;1\right\}\)

\(\left(b\right):\left(2x-3\right)\left(x-4\right)-x+4=0\\ < =>\left(2x-3\right)\left(x-4\right)-\left(x-4\right)=0\\ < =>\left(x-4\right)\left(2x-3-1\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\2x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.=>S=\left\{4;2\right\}\\ \left(d\right):\left(x^2-8\right)\left(x-2020\right)-x+2020=0\\ < =>\left(x^2-8\right)\left(x-2020\right)-\left(x-2020\right)=0\\ < =>\left(x-2020\right)\left(x^2-8-1\right)=0\\ =>\left[{}\begin{matrix}x-2020=0\\x^2-9=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2020\\x=\pm3\end{matrix}\right.=>S=\left\{2020;3;-3\right\}\)

\(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\left(1< x< 2\right)\\ =\dfrac{\sqrt{27}.\sqrt{\left(x-1\right)^2}}{\sqrt{12}}+\dfrac{3}{2}-\left(x-2\right).\dfrac{\sqrt{50}.\sqrt{x^2}}{\sqrt{8}.\sqrt{\left(x-2\right)^2}}\\ =\dfrac{\sqrt{9}.\sqrt{3}.\left|x-1\right|}{\sqrt{4}.\sqrt{3}}+\dfrac{3}{2}-\left(x-2\right).\dfrac{\sqrt{25}.\sqrt{2}.\left|x\right|}{\sqrt{4}.\sqrt{2}.\left|x-2\right|}\\ =\dfrac{3.\left(x-1\right)}{2}+\dfrac{3}{2}-\left(x-2\right).\dfrac{5x}{-2\left(x-2\right)}\\ =\dfrac{3x-3}{2}+\dfrac{3}{2}+\dfrac{5x}{2}\\ =\dfrac{3x-3+3+5x}{2}=\dfrac{8x}{2}=4x\)

\(\left(a\right):A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne\left\{4;9\right\}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

\(\left(b\right):A< 1< =>A-1< 0\\ < =>\dfrac{3}{\sqrt{x}-2}-1< 0\\ < =>\dfrac{3-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\\ < =>\dfrac{5-\sqrt{x}}{\sqrt{x}-2}< 0\\ < =>\dfrac{\sqrt{x}-5}{\sqrt{x}-2}>0\\ =>\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-5>0\\\sqrt{x}-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}-2< 0\end{matrix}\right.\end{matrix}\right.< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>25\\x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 25\\x< 4\end{matrix}\right.\end{matrix}\right.< =>\left[{}\begin{matrix}x>25\\x< 4\end{matrix}\right.\)

\(Kết\ hợp\ ĐK=>0\le x<4\ hoặc\ x>25\ thì\ A<1\)

\(\left(a\right):A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{x-4}\right)\left(x>0;x\ne4\right)\\ =\left(\dfrac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\\ =\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)-2\sqrt{x}}\\ =\dfrac{x+9}{6\sqrt{x}}\)

\(\left(b\right):A< 1< =>A-1< 0\\ < =>\dfrac{x+9}{6\sqrt{x}}-1< 0\\ < =>\dfrac{x+9-6\sqrt{x}}{6\sqrt{x}}< 0\\ < =>\dfrac{\left(\sqrt{x}-3\right)^2}{6\sqrt{x}}< 0\)

\(Ta\ thấy:\) \(\left\{{}\begin{matrix}\left(\sqrt{x}-3\right)^2\ge0\forall x\inĐK\\6\sqrt{x}>0\forall x\inĐK\end{matrix}\right.=>\dfrac{\left(\sqrt{x}-3\right)^2}{6\sqrt{x}}\ge0\forall x\inĐK\)

\(Do\ vậy\ nên\ không\ có\ giá\ trị\ của\ x\ thỏa\ mãn\ A<1\)