a) \(I\left(1;3\right)\) là tâm của đường tròn \(\left(C\right)\)\(;A\left(3;1\right)\in\left(C\right)\)

\(\Rightarrow R=IA=\sqrt{\left(3-1\right)^2+\left(1-3\right)^2}=2\sqrt{2}\)

\(\Rightarrow PTĐT\left(C\right):\left(x-1\right)^2+\left(y-3\right)^2=8\)

b) Tâm \(I\left(-2;0\right);\left(C\right)\) tiếp xúc với \(\left(\Delta\right):2x+y-1=0\)

 \(\Rightarrow R=d\left(I;\left(\Delta\right)\right)=\dfrac{\left|2.\left(-2\right)+0-1\right|}{\sqrt{2^2+1^2}}=\sqrt{5}\)

\(\Rightarrow PTĐT\left(C\right):\left(x+2\right)^2+y^2=5\)

Ta có :

\(AB\perp AC\left(\Delta ABC.vuông.tại.A\right)\)

mà \(SA\perp AB\left(SA\perp\left(ABCD\right)\right)\)

\(\Rightarrow AB\perp\left(SAC\right)\)

mà \(\left\{{}\begin{matrix}AM\subset\left(SAC\right)\left(AM\perp SC\subset\left(SAC\right)\right)\\SA\subset\left(SAC\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}AB\perp AM\\AB\perp SA\end{matrix}\right.\)

\(\Rightarrow AB\perp\left(SAM\right)\left(đpcm\right)\)

(Có thể hiểu rằng \(\left(SAC\right)\equiv\left(SAM\right)\))

\(a=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{20^{20}}+\dfrac{1}{2^{21}}\)

\(\Rightarrow2a=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{20^{19}}+\dfrac{1}{2^{20}}\)

\(\Rightarrow2a-a=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{20^{19}}+\dfrac{1}{2^{20}}\right)-1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{20^{20}}+\dfrac{1}{2^{21}}\)

\(\Rightarrow a=2-\dfrac{1}{2^{21}}=\dfrac{2^{22}-1}{2^{21}}\)

Gọi số học sinh của lớp \(8A\) là \(x\left(x\in N;x\ne0\right)\)

- Học kì \(I\) số học sinh giỏi là \(\dfrac{x}{8}\left(học.sinh\right)\)

- Học kì \(II\) số học sinh giỏi là \(\dfrac{x}{8}+3=\dfrac{20}{100}x=\dfrac{x}{5}\left(học.sinh\right)\)

\(\Rightarrow\dfrac{x}{5}-\dfrac{x}{8}=3\)

\(\Rightarrow\dfrac{3x}{40}=3\)

\(\Rightarrow x=40\left(học.sinh\right)\)

Vậy lớp \(8A\) có \(40\left(học.sinh\right)\)

a) Ta có :

\(CD\perp AD\left(ABCD.là.HCN\right)\)

mà \(CD\perp SA\left(SA\perp\left(ABCD\right)\right)\)

\(\Rightarrow CD\perp\left(SAD\right)\)

b) Ta có : Hình chiếu của \(S\) lên \(\left(ABCD\right)\) là \(A\left(SA\perp\left(ABCD\right)\right)\)

mà Hình chiếu của \(C\) lên \(\left(ABCD\right)\) là \(C\left(C\in\left(ABCD\right)\right)\)

\(\Rightarrow\) Hình chiếu của \(SC\) lên \(\left(ABCD\right)\) là \(AC\)

\(\Rightarrow\left(\widehat{SC;\left(ABCD\right)}\right)=\widehat{SCA}\)

Xét tam giác vuông \(SAC\) (Vuông tại \(A\)); ta có :

\(tan\widehat{SCA}=\dfrac{SA}{AC}\)

mà \(AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+4a^2}=a\sqrt{5}\left(Pitago\right)\)

\(\Rightarrow tan\widehat{SCA}=\dfrac{3a}{a\sqrt{5}}=\dfrac{3\sqrt{5}}{5}\)

\(\Rightarrow\left(\widehat{SC;\left(ABCD\right)}\right)=\widehat{SCA}=arctan\left(\dfrac{3\sqrt{5}}{5}\right)\)

c) Ta có :

\(CD\perp\left(SAD\right)\left(cmt\right)\)

mà \(AH\subset\left(SAD\right)\)

\(\Rightarrow AH\perp CD\)

mà \(AH\perp SD\) (\(AH\) là đường cao \(\Delta SAD\))

\(\Rightarrow AH\perp\left(SCD\right)\)

mà \(SC\subset\left(SCD\right)\)

\(\Rightarrow AH\perp SC\left(đpcm\right)\)

Gọi \(A\left(m;0;0\right);B\left(0;n;0\right);C\left(0;0;p\right)\) lần lượt là các điểm cắt trục \(Ox;Oy;Oz\)

Theo đề bài ta có \(OA;OB;OC\) là một cấp số nhân với công bội \(3\)

\(\Rightarrow OA=m;OB=n=3m;OC=c=3n=9m\)

\(A;B;C\in\left(\alpha\right)\Rightarrow PTTQ\left(\alpha\right):\dfrac{x}{m}+\dfrac{y}{n}+\dfrac{z}{p}=1\)

\(\Rightarrow\left(\alpha\right):\dfrac{x}{m}+\dfrac{y}{3m}+\dfrac{z}{9m}=1\)

\(\Rightarrow\left(\alpha\right):9x+3y+z=9m\)

\(M\left(1;2;3\right)\in\left(\alpha\right)\Rightarrow9.1+3.2+3=9m\Leftrightarrow m=2\)

\(\Rightarrow\left(\alpha\right):9x+3y+z-18=0\)

mà \(\left(\alpha\right):ax+by+z+d=0\)

\(\Rightarrow a=9;b=3;d=-18\)

\(\Rightarrow a+b+d=9+3-18=-6\)

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}=x+y+z\)

\(\Rightarrow\left(\dfrac{x}{a}\right)^2=\left(\dfrac{y}{b}\right)^2=\left(\dfrac{z}{c}\right)^2=\left(x+y+z\right)^2\left(1\right)\)

Ta lại có :

\(\left(\dfrac{x}{a}\right)^2=\left(\dfrac{y}{b}\right)^2=\left(\dfrac{z}{c}\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\left(Đpcm\right)\)

Theo đề bài ta có \(\left(d\right)\cap\left(d'\right)=M\left(1;2;4\right)\)

\(N\left(a;b;c\right)\in\left(d'\right)\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=1+3t'< 0\end{matrix}\right.\) \(\Rightarrow t'< -\dfrac{1}{3}\)

\(\Rightarrow\overrightarrow{MN}=\left(0;0;3t'-3\right)\)

Ta có : \(MN=6\left(cm\right)\)

\(\Leftrightarrow\sqrt{0^2+0^2+\left(3t'-3\right)^2}=6\)

\(\Leftrightarrow\left|3t'-3\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}3t'-3=6\\3t'-3=-6\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}t'=3\left(loại\right)\\t'=-1\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow t'=-1\Rightarrow c=1+3.\left(-1\right)=-2\)

\(\Rightarrow N\left(a;b;c\right)=\left(1;2;-2\right)\)

\(\Rightarrow a-b+c=1-2+\left(-2\right)=-3\)

\(\left(P\right):2x-y+2z+5=0\Rightarrow\overrightarrow{n_P}=\left(2;-1;2\right)\)

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

\(cos\left(\widehat{\left(P\right);\left(Q\right)}\right)=\dfrac{\left|\overrightarrow{n_P}.\overrightarrow{n_Q}\right|}{\left|\overrightarrow{n_P}\right|.\left|\overrightarrow{n_Q}\right|}=\dfrac{\left|2.1+\left(-1\right).\left(-1\right)+2.0\right|}{\sqrt{2^2+\left(-1\right)^2+2^2}.\sqrt{1^2+\left(-1\right)^2}}=\dfrac{3}{3\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left(\widehat{\left(P\right);\left(Q\right)}\right)=45^o\)

\(\Rightarrow A.Sai;B.Đúng\)

\(\dfrac{2}{1}\ne\dfrac{-1}{-1}\ne\dfrac{2}{0}\Rightarrow\left(P\right)\&\left(Q\right)\) không song song với nhau

\(\Rightarrow C.Sai\)

Giả sử \(M\left(0;5;0\right)\in\left(P\right)\Leftrightarrow2.0-5+2.0+5=0\Leftrightarrow0=0\left(đúng\right)\)

\(\Rightarrow D.Đúng\)

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

\(\left(P\right)//\left(Q\right)\Rightarrow\overrightarrow{n_Q}=\left(1;2;3\right)\)

\(\Rightarrow\left(Q\right):x+2y+3z+c=0\)

mà \(\left(Q\right):ax+2y+bz+c=0\)

\(\Rightarrow a=1;b=3\)

\(M\left(1;2;3\right)\in\left(Q\right)\Rightarrow1+2.2+3.3+c=0\Rightarrow c=-14\)

\(\Rightarrow a+b+c=1+3-14=-10\)