\(\overrightarrow{AB}=\left(-6;-3\right)=-3\left(2;1\right)\Rightarrow\) đường thẳng AB nhận \(\left(2;1\right)\) là 1 vtcp

Phương trình tham số đường thẳng AB có dạng: \(\left\{{}\begin{matrix}x=5+2t\\y=4+t\end{matrix}\right.\)

Do M thuộc AB nên tọa độ M có dạng \(M\left(5+2t;4+t\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-2t;-t\right)\\\overrightarrow{MC}=\left(-2-2t;-6-t\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{MA}+\overrightarrow{MC}=\left(-2-4t;-6-2t\right)\)

Đặt \(T=\left|\overrightarrow{MA}+\overrightarrow{MC}\right|=\sqrt{\left(-2-4t\right)^2+\left(-6-2t\right)^2}=\sqrt{20\left(t+1\right)^2+20}\ge\sqrt{20}\)

Dấu "=" xảy ra khi \(t+1=0\Rightarrow t=-1\Rightarrow M\left(3;3\right)\)

1: D thuộc Ox nên D(x;0)

vecto AB=(-3;4)

vecto DC=(-3-x;-1)

Để ABDC là hình thang thì \(\dfrac{-3}{-x-3}=\dfrac{4}{-1}=-4\)

=>3/x+3=4

=>x+3=3/4

=>x=-9/4

2: \(\overrightarrow{MA}=\left(3-x;0\right)\)

vectoMC=(-3-x;-1)

Để |vecto MA+vecto MC| nhỏ nhất thì vecto MA+vecto MC=vecto 0

=>M là trung điểm của AC

=>M(0;-1/2)

a/ \(\overrightarrow{AB}=\left(4;8\right)\Rightarrow\) đường thẳng AB có 1 vtpt là \(\left(2;-1\right)\)

Phương trình AB:

\(2\left(x-3\right)-\left(y-4\right)=0\Leftrightarrow2x-y-2=0\)

A;P;B thẳng hàng \(\Rightarrow P\in AB\Rightarrow P\left(x;2x-2\right)\)

\(\overrightarrow{AP}=\left(x+1;2x+2\right)\Rightarrow AP^2=\left(x+1\right)^2+\left(2x+2\right)^2=5\left(x+1\right)^2\)

\(\Rightarrow5\left(x+1\right)^2=\left(3\sqrt{5}\right)^2\Rightarrow\left(x+1\right)^2=9\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}P\left(2;2\right)\\P\left(-4;-10\right)\end{matrix}\right.\)

Gọi \(M\left(x;0\right)\)

b/ \(\overrightarrow{AM}=\left(x+1;4\right)\Rightarrow MA=\sqrt{\left(x+1\right)^2+4^2}\)

\(\overrightarrow{MB}=\left(3-x;4\right)\Rightarrow MB=\sqrt{\left(3-x\right)^2+4^2}\)

\(T=MA+MB=\sqrt{\left(x+1\right)^2+4^2}+\sqrt{\left(3-x\right)^2+4^2}\)

Áp dụng BĐT Mincopxki:

\(T\ge\sqrt{\left(x+1+3-x\right)^2+\left(4+4\right)^2}=4\sqrt{5}\)

\(T_{min}=4\sqrt{5}\) khi \(x+1=3-x\Rightarrow x=1\Rightarrow M\left(1;0\right)\)

c/ Tương tự như câu b:

\(MB+MC=\sqrt{\left(3-x\right)^2+4^2}+\sqrt{\left(x-2\right)^2+5^2}\)

\(MB+MC\ge\sqrt{\left(3-x+x-2\right)^2+\left(4+5\right)^2}=\sqrt{82}\)

Dấu "=" xảy ra khi \(\frac{3-x}{4}=\frac{x-2}{5}\Rightarrow x=\frac{23}{9}\Rightarrow M\left(\frac{23}{9};0\right)\)

minh ko bik

Gọi `M(x;3/2x+5/2)`

Ta có:`|\vec{MA}-2\vec{MB}|`

`=|(4-x;7-3/2x-5/2)-2(2-x;1-3/2x-5/2)|`

`=|(x;3/2x+17/2)|`

`=\sqrt{x^2+(3/2x+17/2)^2}`

`=\sqrt{x^2+9/4x^2+51/2x+289/4}`

`=\sqrt{13/4x^2+51/2x+289/4}`

`=\sqrt{(\sqrt{13}/2 x+[51\sqrt{13}]/26)^2+289/13} >= [17\sqrt{13}]/13`

Dấu "`=`" xảy ra `<=>\sqrt{13}/2x+[51\sqrt{13}]/26=0<=>x=-51/13`

   `=>M(-51/13;-44/13)`