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Gọi \(M\left(x;0\right)\Rightarrow\overrightarrow{MA}\left(2-x;5\right)\) ; \(\overrightarrow{MB}=\left(-1-x;8\right)\); \(\overrightarrow{MC}=\left(4-x;-3\right)\)
a/ \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(5-3x;10\right)\)
\(\Rightarrow T=\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\sqrt{\left(5-3x\right)^2+10^2}\ge10\)
\(T_{min}=10\) khi \(5-3x=0\Rightarrow x=\frac{5}{3}\Rightarrow M\left(\frac{5}{3};0\right)\)
b/ \(2\overrightarrow{MA}-\overrightarrow{MB}+3\overrightarrow{MC}=\left(17-4x;-7\right)\)
\(\Rightarrow A=\left|2\overrightarrow{MA}-\overrightarrow{MB}+3\overrightarrow{MC}\right|=\sqrt{\left(17-4x\right)^2+\left(-7\right)^2}\ge7\)
\(A_{min}=7\) khi \(17-4x=0\Rightarrow x=\frac{17}{4}\Rightarrow M\left(\frac{17}{4};0\right)\)
Gọi \(M\left(x;0\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-1-x;4\right)\\\overrightarrow{MB}=\left(1-x;-2\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{MA}+2\overrightarrow{MB}=\left(1-3x;0\right)\)
\(\Rightarrow\left|\overrightarrow{MA}+2\overrightarrow{MB}\right|=\sqrt{\left(1-3x\right)^2}\ge0\)
Dấu "=" xảy ra khi \(x=\frac{1}{3}\Rightarrow M\left(\frac{1}{3};0\right)\)
Gọi \(P\left(0;y\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{PA}=\left(-1;4-y\right)\\\overrightarrow{PB}=\left(1;-2-y\right)\\\overrightarrow{PC}=\left(3;4-y\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{PA}+2\overrightarrow{PB}-4\overrightarrow{PC}=\left(-11;5y-16\right)\)
\(\Rightarrow\left|\overrightarrow{PA}+\overrightarrow{PB}-4\overrightarrow{PC}\right|=\sqrt{11^2+\left(5y-16\right)^2}\ge11\)
Dấu "=" xảy ra khi \(5y-16=0\Rightarrow y=\frac{16}{5}\Rightarrow P\left(0;\frac{16}{5}\right)\)
Câu 1:
Vì \(\overrightarrow{BA}\uparrow\uparrow\overrightarrow{CD}\) và \(BA=\frac{1}{3}CD\Rightarrow \overrightarrow{BA}=\frac{1}{3}\overrightarrow{CD}\)
Để $B,M,D$ thẳng hàng \(\Leftrightarrow \exists k\in\mathbb{R}|\overrightarrow{BM}=k\overrightarrow{MD}\)
\(\Leftrightarrow \overrightarrow{BA}+\overrightarrow{AM}=k\overrightarrow{MD}\)
\(\Leftrightarrow \frac{1}{3}\overrightarrow{CD}+x\overrightarrow{MC}=k\overrightarrow{MD}\)
\(\Leftrightarrow \frac{1}{3}(\overrightarrow{MC}+\overrightarrow{CD})+(x-\frac{1}{3})\overrightarrow{MC}=k\overrightarrow{MD}\)
\(\Leftrightarrow \frac{1}{3}\overrightarrow{MD}+(x-\frac{1}{3})\overrightarrow{MC}=k\overrightarrow{MD}\)
\(\Leftrightarrow (x-\frac{1}{3})\overrightarrow{MC}=(k-\frac{1}{3})\overrightarrow{MD}\)
Vì \(\overrightarrow{MC}; \overrightarrow{MD}\) không phải 2 vecto cùng phương nên điều trên chỉ xảy ra khi \(x-\frac{1}{3}=k-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)
Bài 2:
Lấy điểm $I(a,b)$ sao cho \(\overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC}=\overrightarrow{0}\)
\(\Leftrightarrow (1-a, 1-b)-2(4-a, 3-b)+3(2-a, -2-b)=(0,0)\)
\(\Leftrightarrow (-1-2a,-11-2b)=(0,0)\Rightarrow a=-\frac{1}{2}; b=\frac{-11}{2}\)
Vậy \(I(-\frac{1}{2}; -\frac{11}{2})\)
Ta có:
\(|\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}|=|\overrightarrow{MI}+\overrightarrow{IA}-2(\overrightarrow{MI}+\overrightarrow{IB})+3(\overrightarrow{MI}+\overrightarrow{IC})|\)
\(|2\overrightarrow{MI}+(\overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC})|=2|\overrightarrow{MI}|\)
Để \(|\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}|\) min thì \(|\overrightarrow{MI}|\) min. Điều này xảy ra khi $M$ là hình chiếu của $I$ trên $Ox$
Do đó \(M=(-\frac{1}{2};0)\)
gọi M có tọa độ là (x;y) do M thuộc Ox=> tọa ddoooj M là (x;0)
ta có : \(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|\left(-2-X;5\right)+\left(3-X;-1\right)+\left(7-X;1\right)\right|\)
=\(\left|\sqrt{\left(-2-X\right)^2+5^2}+\sqrt{\left(3-X\right)^2+1}+\sqrt{\left(7-X\right)^2+1}\right|\)
=> BẠN TÌ gtnn CÁI TRONG LÀ ĐC
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 \(\frac{3-x}{4}=\frac{x-2}{5}\Rightarrow x=\frac{23}{9}\Rightarrow M\left(\frac{23}{9};0\right)\)