Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. a) D = [1;4] \{2;3}
b) D = (0;+∞)
2.
\(2\overrightarrow{a}\)= (2;4) và \(3\overrightarrow{b}\) = (9;12)
⇒ \(2\overrightarrow{a}\) + \(3\overrightarrow{b}\) = (2+9; 4+12)
⇔ (11; 16)
Vậy \(\overrightarrow{m}\) = (11;16)
b) \(\left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|\) khi vectơ a và vectơ b cùng hướng
\(A^2=\left|3a+5b\right|^2=9a^2+25b^2+30ab=9.1+25.1+30.3=124\)
\(\Rightarrow A=2\sqrt{31}\)
Hok nhanh phết, chưa j đã đến phần toạ độ vecto r
1/ \(\overrightarrow{MB}=\left(x_B-x_M;y_B-y_M\right)=\left(2-x_M;3-y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}=\left(4-2x_M;6-2y_M\right)\)
\(\overrightarrow{3MC}=\left(3x_C-3x_M;3y_C-3y_M\right)=\left(-3-3x_M;6-3y_M\right)\)
\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(4-2x_M-3-3x_M;6-2y_M+6-3y_M\right)=0\)
\(\Leftrightarrow\left(1-5x_M;12-5y_M\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-5x_M=0\\12-5y_M=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_M=\frac{1}{5}\\y_M=\frac{12}{5}\end{matrix}\right.\Rightarrow M\left(\frac{1}{5};\frac{12}{5}\right)\)
2/ \(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2+9;4+12\right)=\left(11;16\right)\)
3/ \(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)=\left(-5-3;4+2\right)=\left(-8;6\right)\)
\(\overrightarrow{AC}=\left(x_C-x_A;y_C-y_A\right)=\left(\frac{1}{3}-3;0+2\right)=\left(-\frac{8}{3};2\right)\)
\(\Rightarrow x=\frac{\overrightarrow{AB}}{\overrightarrow{AC}}=\frac{\left(-8;6\right)}{\left(-\frac{8}{3};2\right)}=3\)
Câu 4 tương tự
Câu 5 vt lại đề bài nhé bn, nghe nó vô lý sao á, m,n ở đâu ra vậy, cả A,B,C nx
\(\left(a+2b\right)^2=28\Leftrightarrow a^2+4b^2+4ab=28\)
\(\Rightarrow ab=\frac{28-4^2-4.3^2}{4}=-6\)
\(\Rightarrow cos\left(a;b\right)=-\frac{6}{4.3}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)
\(u.v=0\Leftrightarrow\left(2a+3b\right)\left(-15a+14b\right)=0\)
\(\Leftrightarrow-30a^2+42b^2-17ab=0\)
\(\Leftrightarrow ab=\frac{-30.4^2+42.3^2}{17}=-6\)
\(\Rightarrow cos\left(a;b\right)=\frac{ab}{\left|a\right|\left|b\right|}=-\frac{6}{12}=-\frac{1}{2}\Rightarrow\left(a;b\right)=120^0\)
a, \(\left|\overrightarrow{a}\right|=\sqrt{3^2+4^2}=5\)
\(\left|\overrightarrow{b}\right|=\sqrt{7^2+1^2}=5\sqrt{2}\)
\(cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{3.7+4.1}{5.5\sqrt{2}}=\frac{\sqrt{2}}{2}\) \(\Rightarrow\left(\overrightarrow{a},\overrightarrow{b}\right)=45^0\)
b, Gọi \(\overrightarrow{c}\left(x;y\right)\)
\(\overrightarrow{a}.\overrightarrow{c}=15\)
\(\Leftrightarrow3x+4y=15\)
\(\overrightarrow{b}.\overrightarrow{c}=10\)
\(\Leftrightarrow7x+y=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
\(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2;4\right)+\left(9;12\right)=\left(11;16\right)\)