$\displaystyle \begin{array}{{>{\displaystyle}l}} Câu\ 1:\ \\ Ta\ có\ :\ 1,25=\frac{5}{4} \ \\ Ta\ có\ :\ \frac{5}{4} >\frac{-5}{4} \ và\ \frac{5}{4} >\frac{-4}{3} \ \\ Ta\ có\ :\ \frac{-5}{4} =\frac{-15}{12} \ và\ \frac{-4}{3} =\frac{-16}{12} \ \\ Ta\ có\ :\ -15 >-16\ \\ \rightarrow \frac{-5}{4} >\frac{-4}{3} \ \\ Do\ đó\ :\ \frac{5}{4} >\frac{-5}{4} >\frac{-4}{3} \ \\ hay\ 1,25 >\frac{-5}{4} >\frac{-4}{3} \ \\ Câu\ 2:\ \frac{-8}{18} -\frac{15}{27} =\frac{-4}{9} -\frac{5}{9} =\frac{-9}{9} =-1\\ Câu\ 3:\ \left( 0.2-\frac{1}{3} x\right)\left( 0.2+\frac{1}{3} x\right) \ \\ =0.2^{2} -\frac{1}{9} x^{2} =0.04-\frac{1}{9} x^{2}\\ Câu\ 4:\ \left(\frac{1}{2} x-0.5\right)^{2}\\ =\frac{1}{4} x^{2} -2.\frac{1}{2} .\frac{1}{2} x+0.25\\ =\frac{1}{4} x^{2} -\frac{1}{2} x+0.25\\ Câu\ 5:\ \left( 2x-\frac{1}{3}\right)^{2}\\ =8x^{3} -3.( 2x)^{2} .\frac{1}{3} +3.2x.\frac{1}{9} -\frac{1}{27}\\ =8x^{3} -4x^{2} +\frac{2}{9} x-\frac{1}{27}\\ Câu\ 6:\ x^{2}( x-y) -( x-y) =( x-y)\left( x^{2} -1\right)\\ =( x-y)( x-1)( x+1) \ \\ Câu\ 7\ :\ Nhiều\ nhất\ 3\ góc\ vuông\ \\ Câu\ 8\ :\ C\ \\ Câu\ 9\ :\ có\ 2\ góc\ kề\ 1\ cạnh\ =90\ \\ \\ \\ \\ \end{array}$
