Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 41.D\\ 42.C\\ 43.D\\ 44.B\\ 45.B\\ 46.D\\ 47.B\\ 48.B\\ 49.A\\ 50.D \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 41.D\\ MD+MB=MA+MC\\ \Rightarrow MD=MA+MC-MB\\ 42.C\\ Ox\cap ( d) =A\Rightarrow A\left(\frac{5-m}{2m-3} ;0\right)\\ Oy\cap ( d) =B\Rightarrow B( 0;m-5)\\ \Delta OAB\ cân\ tại\ O\Rightarrow |\frac{5-m}{2m-3} |=|m-5|\\ \Leftrightarrow ( 5-m)^{2} =( m-5)^{2}( 2m-3)^{2}\\ TH1:5-m=0\\ \Leftrightarrow m=5\\ TH2:\ ( 2m-3)^{2} =1\\ \Leftrightarrow 4m^{2} -12m+8=0\\ \Leftrightarrow m=2;\ m=1\\ \Rightarrow S=1+2+5=8\\ 43.D\\ Gọi\ I\ là\ tâm\ đường\ tròn\ ngoại\ tiếp.\\ \widehat{BAI} =\frac{1}{2}\hat{A} =60^{o}\\ \Delta ABI\ có\ BI=IA=R;\ \widehat{BAI} =60^{o}\\ \Rightarrow \Delta ABI\ là\ tam\ giác\ đều\\ \Rightarrow R=3\\ \Rightarrow Độ\ dài\ đường\ tròn:\ 2\pi R=6\pi \\ 44.B\\ ( I) \Leftrightarrow ( x;y) =( 3;1)\\ Để\ ( I) \ và\ ( II) \ tương\ đương\ nhau\ \Leftrightarrow ( x;y) =( 3;1) \ là\ nghiệm\ của( II)\\ \Rightarrow \{_{9a+2b=10}^{3a+b=6} \Leftrightarrow ( a;b) =\left( -\frac{2}{3} ;8\right)\\ \Rightarrow T=6.\left( -\frac{2}{3}\right) +8=4\\ 45.B\\ Xét\ trường\ hợp\ OO'=5+3=8\\ Có\ OE//O'F\\ Kẻ\ EH\bot O'F\\ \Rightarrow HF=5-3=2\\ \Rightarrow EF=\sqrt{8^{2} +2^{2}} =2\sqrt{17}\\ 46.D\\ ĐL\ cos:\ AC^{2} =AB^{2} +BC^{2} -2AB.BC^{2} .cosB\\ \Rightarrow AC=\sqrt{8^{2} +6^{2} -2.6.8.cos60^{o}} =2\sqrt{13}\\ 47.B\\ P-\frac{48}{13} =\frac{3x-16}{\left(\sqrt{x} -1\right)^{2}} -\frac{48}{13} =\frac{3x-16}{x-2\sqrt{x} +1} -\frac{48}{13}\\ =\frac{3x-16-\frac{48}{13} x+\frac{96}{13}\sqrt{x} -\frac{48}{13}}{x-2\sqrt{x} +1} =\frac{-\frac{9}{13} x+\frac{96}{13}\sqrt{x} -\frac{256}{13}}{x-2\sqrt{x} +1} \leqslant 0\\ \Rightarrow P\leqslant \frac{48}{13}\\ \Rightarrow T=48+13=61\\ 48.B\\ \frac{1}{3^{2}} +\frac{1}{n^{2}} +\frac{1}{( n+3)^{2}} =\left(\frac{1}{3} +\frac{1}{n} -\frac{1}{n+1}\right)^{2}\\ \Rightarrow P=\sqrt{\left(\frac{1}{3} +1-\frac{1}{4}\right)^{2}} +\sqrt{\left(\frac{1}{3} +\frac{1}{4} -\frac{1}{7}\right)^{2}} +...+\sqrt{\left(\frac{1}{3} +\frac{1}{592} -\frac{1}{595}\right)^{2}}\\ P=\frac{1}{3} +1-\frac{1}{4} +\frac{1}{3} +\frac{1}{4} -\frac{1}{7} +...+\frac{1}{3} +\frac{1}{592} -\frac{1}{595}\\ P=\frac{1}{3} .\left(\frac{592-1}{3} +1\right) +1-\frac{1}{595} =\frac{39864}{595}\\ \Rightarrow Q=39864-66.595=594\\ 49.A\\ a+b+c-21=2\left(\sqrt{a-7} +\sqrt{b-8} +\sqrt{c-9}\right)\\ \Leftrightarrow \left( a-7-2\sqrt{a-7} +1\right) +\left( b-8-2\sqrt{b-8} +1\right) +\left( c-9-2\sqrt{c-9} +1\right) =0\\ \Leftrightarrow \left(\sqrt{a-7} -1\right)^{2} +\left(\sqrt{b-8} -1\right)^{2} +\left(\sqrt{c-9} -1\right)^{2} =0\\ \Leftrightarrow \sqrt{a-7} -1=\sqrt{b-8} -1=\sqrt{c-9} -1=0\\ \Leftrightarrow a=8;\ b=9;\ c=10\\ \Rightarrow S=16\\ 50.D\\ Để\ PT\ có\ 2\ nghiệm\ phân\ biệt\ \Leftrightarrow m\neq 0;\ \Delta '=( m-4)^{2} -m( m-4) \geqslant 0\\ \Leftrightarrow m\neq 0;\ -4m+16\geqslant 0\\ \Rightarrow m\neq 0;\ m\leqslant 4\\ mà\ -7\leqslant m\leqslant 7\\ \Rightarrow m=\{-7;-6;-5;-3;-2;-1;1;2;3;4\}\\ \end{array}$
