\(\left(a\right):A=x^2y+x.\left(-xy\right)-3y\\ =x^2y-x^2y-3y=-3y\)

\(Với\ y=-1=>A=-3.(-1)=3\)

\(\left(b\right):B=3x^2y^2-6x^2y+9xy\left(-xy\right)+1\\ =3x^2y^2-6x^2y-9x^2y^2+1\\ =-6x^2y^2-6x^2y+1=-6x^2y\left(y+1\right)+1\)

\(Với\ x=2;y=-1=>B=-6.2^2.(-1).(-1+1)+1=-6.2^2.(-1).0+1=0+1=1\)\(\left(c\right):C=x^2+y^2-2x^2-4y^2=-x^2-3y^2\)

\(Với\ x=-1;y=-2=>C=-(-1)^2-3.(-2)^2=-1-3.4=-1-12=-13\)

\(\left(x+2\right)\left(\dfrac{120}{x}+1\right)=160\left(x\ne0\right)\\ < =>120+x+\dfrac{240}{x}+2=160\\ < =>x+\dfrac{240}{x}-38=0\\ < =>\dfrac{x^2+240-38x}{x}=0\\ < =>x^2-38x+240=0\\ < =>\left(x^2-38x+19^2\right)-121=0\\ < =>\left(x-19\right)^2=121=\left(\pm11\right)^2\\ =>\left[{}\begin{matrix}x-19=11\\x-19=-11\end{matrix}\right.< =>\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\left(TMDK\right)=>S=\left\{30;8\right\}\)

\(\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{15}{-7}.\left(-32\right)\\ =-\dfrac{7.5.15.32}{15.8.7}=-\dfrac{7.5.15.8.4}{15.8.7}\\ =-\dfrac{15.8.7}{15.8.7}.5.4=\left(-1\right).20=-20\)

\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\\ =>\left|x+\dfrac{3}{4}\right|=\dfrac{1}{2}\\ =>\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{2}\\x+\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(\left(a\right):A=\left(\dfrac{2}{x^2-2x}+\dfrac{1}{x-2}\right):\dfrac{x+2}{x^2-4x+4}\left(x\ne\left\{0;\pm2\right\}\right)\\ =\left(\dfrac{2}{x\left(x-2\right)}+\dfrac{1}{x-2}\right):\dfrac{x+2}{\left(x-2\right)^2}\\ =\dfrac{2+x}{x\left(x-2\right)}.\dfrac{\left(x-2\right)^2}{x+2}=\dfrac{x-2}{x}\)

\(\left(b\right):x=\dfrac{1}{9}\left(TMDK\right)=>A=\dfrac{\dfrac{1}{9}-2}{\dfrac{1}{9}}=-\dfrac{17}{9}:\dfrac{1}{9}=-\dfrac{17}{9}.9=-17\)

\(\left(c\right):A=\dfrac{x-2}{x}=\dfrac{x}{x}-\dfrac{2}{x}=1-\dfrac{2}{x}\\ A\in Z=>\dfrac{2}{x}\in Z\\ =>2⋮x=>x\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ Do:x\ne\left\{0;\pm2\right\}=>x\in\left\{\pm1\right\}\)

\(Vậy\ các\ giá\ trị\ nguyên\ của\ x\ thỏa\ mãn\ A\ nguyên\ là :1;-1\)

\(\left(a\right):\left(\dfrac{-1}{2}\right)^2.\left(\dfrac{2}{5}\right)^2=\left(\dfrac{-1}{2}.\dfrac{2}{5}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\ \left(b\right):\left(\dfrac{5}{2}\right)^3.\left(\dfrac{4}{5}\right)^3=\left(\dfrac{5}{2}.\dfrac{4}{5}\right)^3=2^3=8\)

\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\\ < =>x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\\ < =>x^3-25x-x^3-8=3\\ < =>-25x=11\\ < =>x=-\dfrac{11}{25}\)

\(3^{x+2}+3^x=810\\ =>3^x.\left(3^2+1\right)=810\\ =>3^x=\dfrac{810}{3^2+1}=\dfrac{810}{10}=81=3^4\\ =>x=4\)

\(7\sqrt{2}=\sqrt{49}.\sqrt{2}=\sqrt{98}\\ 2\sqrt{8}=\sqrt{4}.\sqrt{8}=\sqrt{32}\\ 5\sqrt{2}=\sqrt{25}.\sqrt{2}=\sqrt{50}\)

\(Vì\ :\ \)\(\sqrt{98}>\sqrt{50}>\sqrt{32}>\sqrt{28}\)

\(=>7\sqrt{2}>5\sqrt{2}>2\sqrt{8}>\sqrt{28}\)

\(=>\ Sắp\ xếp\ theo\ thứ\ tự\ giảm\ dần :\) \(7\sqrt{2};5\sqrt{2};2\sqrt{8};\sqrt{28}\)

\(Ta\ thấy :\)

\(-\dfrac{1}{10}< 0;\dfrac{1}{1000}>0\\ =>-\dfrac{1}{10}< 0< \dfrac{1}{1000}\)

\(Vậy :\) \(-\dfrac{1}{10}< \dfrac{1}{1000}\)