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Câu 2: Ta có: $\widehat{MAN}=\widehat{MBA}=90^0$ $\widehat{ANB}=\widehat{ABN}=45^0$ $\Rightarrow$ tam giác ANB cân tại N $\Rightarrow NA=NB=1(m)$ $\Rightarrow MA=NB=1(m)$ $\Rightarrow MN=AB=15(m)$ $\Rightarrow AM=BN=1(m)$ $\Rightarrow MN=AB=15(m)$ $\Rightarrow AN=BN=15(m)$ Xét tam giác CAN có: $\widehat{CAN}=60^0$ $\widehat{ANC}=90^0$ $\Rightarrow \widehat{ACN}=30^0$ $\Rightarrow AC=2CN$ Áp dụng định lý Pytago vào tam giác CAN ta có: $AC^2=AN^2+CN^2$ $\Rightarrow 4CN^2=15^2+CN^2$ $\Rightarrow 3CN^2=15^2$ $\Rightarrow CN^2=\frac{15^2}{3}=75$ $\Rightarrow CN=5\sqrt{3}(m)$ $\Rightarrow AC=10\sqrt{3}(m)$ $\Rightarrow CD=AC+AD=10\sqrt{3}+1\approx 18(m)$ Câu 3: $xy-3x+2y=0$ $\Leftrightarrow x(y-3)+2(y-3)=6$ $\Leftrightarrow (x+2)(y-3)=6$ Ta có: $6=(-1)\times (-6);(-2)\times (-3);1\times 6;2\times 3$ $\Rightarrow x=-3;y=-3$ $x=-4;y=-2$ $x=-1;y=4$ $x=0;y=5$