a) Đáy \(ABCD\) là hình vuông cạnh \(2\sqrt{2}\)​ và \(SA=4.\) Đặt hệ trục tọa độ với:

\(A\left(0;0;0\right):B\left(2\sqrt{2};0;0\right);C\left(2\sqrt{2};2\sqrt{2};0\right);D\left(0;2\sqrt{2};0\right);S\left(\sqrt{2};\sqrt{2};2\sqrt{3}\right)\)

Xét \(\left(SAC\right):\left\{{}\begin{matrix}\overrightarrow{SA}=\left(-\sqrt{2};-\sqrt{2};-2\sqrt{3}\right)\\\overrightarrow{SC}=\left(\sqrt{2};\sqrt{2};-2\sqrt{3}\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{n_1}=\left[\overrightarrow{SA}.\overrightarrow{SC}\right]=\left(4\sqrt{6};-4\sqrt{6};0\right)=\left(1;-1;0\right)\)

Xét \(\left(SCD\right):\left\{{}\begin{matrix}\overrightarrow{SD}=\left(-\sqrt{2};\sqrt{2};-2\sqrt{3}\right)\\\overrightarrow{SC}=\left(\sqrt{2};\sqrt{2};-2\sqrt{3}\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{n_2}=\left[\overrightarrow{SC}.\overrightarrow{SD}\right]=\left(0;4\sqrt{6};4\right)=\left(0;\sqrt{6};1\right)\)

\(cos\alpha=cos\left(\widehat{\left(SAC\right);\left(SCD\right)}\right)\dfrac{\left|1.0+\left(-1\right).\sqrt{6}+0.1\right|}{\sqrt{1^2+\left(-1\right)^2}.\sqrt{\left(\sqrt{6}\right)^2+1^2}}=\dfrac{\sqrt{6}}{\sqrt{14}}=\dfrac{\sqrt{21}}{7}\)

\(\Rightarrow\alpha\approx49^0\)

b) \(\left(SCD\right):\sqrt{6}\left(y-2\sqrt{2}\right)+\left(z-0\right)=0\)

\(\Rightarrow\left(SCD\right):\sqrt{6}y+z-4\sqrt{3}=0\)

\(d\left(B;\left(SCD\right)\right)=\dfrac{\left|\sqrt{6}.0+0-4\sqrt{3}\right|}{\sqrt{\left(\sqrt{6}\right)^2+1^2}}=\dfrac{4\sqrt{3}}{\sqrt{7}}=\dfrac{4\sqrt{21}}{7}\)

Dùng điều kiện \(P=a^2+b^2-c>0\) sẽ giải nhanh hơn

a) \(\left(m+1\right)^2+\left(-m\right)^2-3m^2+2>0\Leftrightarrow-1< m< 3\)

b) \(\left(m-3\right)^2+\left(-2m\right)^2+m^2-5m-4>0\Leftrightarrow m< \dfrac{5}{6}\cup m>1\)

c) \(m^2+\left(m^2-1\right)^2-m^4+2m^2+4m+3>0\Leftrightarrow\left(m+2\right)^2>0\Leftrightarrow m\ne-2\)

Đã giải rồi, bạn xem phần dưới

Sửa lại đề bài :

\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)

\(\Rightarrow x+y+z=\dfrac{1}{2}\)

\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\Rightarrow\dfrac{x}{\dfrac{1}{2}-x+1}=\dfrac{1}{2}\Rightarrow2x=\dfrac{3}{2}-x\Rightarrow x=\dfrac{1}{2}\)

\(\dfrac{y}{z+x+1}=\dfrac{1}{2}\Rightarrow\dfrac{y}{\dfrac{1}{2}-y+1}=\dfrac{1}{2}\Rightarrow2y=\dfrac{3}{2}-y\Rightarrow y=\dfrac{1}{2}\)

\(x+y+z=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}-\left(x+y\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)=-\dfrac{1}{2}\)

Vậy \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{2};-\dfrac{1}{2}\right)\)

Bài 3 : Theo đề tôi thấy \(A\left(0;3\right);B\left(0;4\right);C\left(4;7\right)\)

Vị trí đèn để chiếu sáng toàn bộ công viên nên vị trí của nó là tâm \(I\left(x;y\right)\) của đường tròn \(\left(C\right)\) ngoại tiếp tam giác \(ABC\)

\(\Rightarrow\left\{{}\begin{matrix}IA^2=IB^2\left(1\right)\\IA^2=IC^2\left(2\right)\end{matrix}\right.\)

\(IA^2=x^2+\left(y-3\right)^2\)

\(IB^2=x^2+\left(y-4\right)^2\)

\(IC^2=\left(x-4\right)^2+\left(y-7\right)^2\)

\(\left(1\right)\Rightarrow\left(y-3\right)^2=\left(y-4\right)^2\)

\(\Leftrightarrow y^2-6y+9=y^2-8y+16\)

\(\Leftrightarrow y=\dfrac{7}{2}\)

\(\left(2\right)\Rightarrow x^2+\left(y-3\right)^2=\left(x-4\right)^2+\left(y-7\right)^2\)

\(\Leftrightarrow8x=-8y+56\)

\(\Leftrightarrow x=\dfrac{-8y+56}{8}=\dfrac{-8.\dfrac{7}{2}+56}{8}=\dfrac{28}{8}=\dfrac{7}{2}\left(thay.y=\dfrac{7}{2}\right)\)

Vậy vị trí đặt cây đèn là tại điểm \(I\left(\dfrac{7}{2};\dfrac{7}{2}\right)\) để thỏa đề bài

a) Kiểm tra \(M\left(-3;1\right)\) có thuộc \(\left(C\right)\) hay không? Ta thay \(M\) vào \(\left(C\right)\)

\(\left(-3-1\right)^2+\left(1+2\right)^2=\left(-4\right)^2+3^2=25=R^2\left(Đúng\right)\)

\(\Rightarrow M\left(-3;1\right)\in\left(C\right)\)

\(\overrightarrow{IM}=\left(-4;3\right)\)

mà \(\overrightarrow{IM}\perp\left(\Delta\right)\) là tiếp tuyến của \(\left(C\right)\)

\(\Rightarrow\overrightarrow{u_{IM}}=\overrightarrow{n_{\Delta}}=\left(-4;3\right)\)

\(\Rightarrow\left(\Delta\right):-4\left(x+3\right)+3\left(y-1\right)=0\)

\(\Rightarrow\left(\Delta\right):4x-3y+15=0\)

b) \(M\left(5;y\right)\in\left(C\right)\Leftrightarrow\left(5-1\right)^2+\left(y+2\right)^2=25\)

\(\Leftrightarrow\left(y+2\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}y+2=3\\y+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\) \(\Rightarrow M_1\left(5;1\right)\cup M_2\left(5;-5\right)\)

\(TH1:M_1\left(5;1\right)\)

\(\Rightarrow\left(\Delta\right):\left(5-1\right)\left(x-1\right)+\left(1+2\right)\left(y+2\right)=25\)

\(\Rightarrow\left(\Delta\right):4x+3y-23=0\)

\(TH2:M_2\left(5;-5\right)\)

\(\Rightarrow\left(\Delta\right):\left(5-1\right)\left(x-1\right)+\left(-5+2\right)\left(y+2\right)=25\)

\(\Rightarrow\left(\Delta\right):4x-3y-35=0\)

c) \(M\left(x;2\right)\in\left(C\right)\Leftrightarrow\left(x-1\right)^2+\left(2+2\right)^2=25\)

\(\Leftrightarrow\left(x-1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) \(\Rightarrow M_1\left(4;2\right)\cup M_2\left(-2;2\right)\)

Tương tự như câu b ta được

\(TH1:M_1\left(4;2\right)\Rightarrow\left(\Delta\right):3x+4y-20=0\)

\(TH2:M_1\left(-2;2\right)\Rightarrow\left(\Delta\right):3x-4y+14=0\)

\(\left(Q\right):x+3y-z-2=0\Rightarrow\overrightarrow{n_Q}=\left(1;3;-1\right)\)

\(\left(T\right):2x+6y-2z+15=0\Rightarrow\overrightarrow{n_T}=\left(2;6;-2\right)\)

\(\Rightarrow\overrightarrow{n_T}=2\overrightarrow{n_Q}\)

\(\Rightarrow\left(Q\right)//\left(T\right)\)

\(\Rightarrow\) Chiều rộng của bức tường \(\left(Q\right)\) là :

\(d\left(\left(Q\right);\left(T\right)\right)=\dfrac{\left|\dfrac{15}{2}+2\right|}{\sqrt{1^2+3^2+\left(-1\right)^2}}=\dfrac{\dfrac{19}{2}}{\sqrt{11}}=\dfrac{19\sqrt{11}}{22}\approx2,86\left(m\right)\)

\(13.B\)

\(14.D\)

Ta có :

\(DM\cap AB=O\)

mà \(O\in AB\subset\left(SAB\right)\)

\(\Rightarrow DM\cap\left(SAB\right)=O\)

mà \(AD\perp AB;BM\perp AB\left(t/c.hình.vuông\right)\)

\(AB\subset\left(SAB\right)\)

\(\Rightarrow tan\widehat{DM;\left(SAB\right)}=tan\widehat{DOA}=tan\widehat{MDC}=\dfrac{MC}{CD}=\dfrac{\dfrac{a}{2}}{a}=\dfrac{1}{2}\)

Ta thấy \(\left\{{}\begin{matrix}AB\subset\left(SAB\right)\\SB\subset\left(SAB\right)\end{matrix}\right.\)

mà \(AB\) cắt \(SB\) tại \(B\)

\(\Rightarrow d\left(AB;SB\right)=0\)

(Bạn xem lại câu hỏi đề bài)