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18.
\(\left|\overrightarrow{CB}-\overrightarrow{CA}\right|=\left|\overrightarrow{CB}+\overrightarrow{AC}\right|=\left|\overrightarrow{AB}\right|=AB=a\)
19.
\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|=AC=\sqrt{AB^2+AD^2}=5\)
20.
Gọi M là trung điểm AC \(\Rightarrow BM=\dfrac{a\sqrt{3}}{2}\) (trung tuyến tam giác đều)
\(\left|\overrightarrow{AB}-\overrightarrow{BC}\right|=\left|-\overrightarrow{BA}-\overrightarrow{BC}\right|=\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=\left|2\overrightarrow{BM}\right|=2BM=2.\dfrac{a\sqrt{3}}{2}=a\sqrt{3}\)
21.
\(\overrightarrow{OB}=\overrightarrow{DO}\Rightarrow\left|\overrightarrow{OA}+\overrightarrow{OB}\right|=\left|\overrightarrow{OA}+\overrightarrow{DO}\right|=\left|\overrightarrow{DA}\right|=AD=a\)
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3.
TH1: \(m=0,pt\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
TH2: \(m\ne0\)
a, Phương trình có hai nghiệm trái dấu khi \(m\left(4m-1\right)< 0\Leftrightarrow0< m< \dfrac{1}{4}\)
b, Phương trình có hai nghiệm phân biệt khi \(\Delta'=-3m^2-m+1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m< \dfrac{-1-\sqrt{13}}{6}\\m>\dfrac{-1+\sqrt{13}}{6}\end{matrix}\right.\)
c, Phương trình có hai nghiệm dương khi:
\(\left\{{}\begin{matrix}\Delta'>0\\x_1x_2>0\\x_1+x_2>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m< -\dfrac{1+\sqrt{13}}{6}\\m>1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
\(a,\Leftrightarrow\Delta'=\left(1-m\right)^2-\left(m^2-m\right)>0\\ \Leftrightarrow m^2-2m+1-m^2+m>0\\ \Leftrightarrow1-m>0\Leftrightarrow m< 1\\ b,\text{Áp dụng Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(1-m\right)\\x_1x_2=m^2-m\end{matrix}\right.\\ \left(2x_1-1\right)\left(2x_2-1\right)-x_1x_2=1\\ \Leftrightarrow2x_1x_2-2\left(x_1+x_2\right)+1-x_1x_2=1\\ \Leftrightarrow x_1x_2-2\left(x_1+x_2\right)=0\\ \Leftrightarrow m^2-m-4\left(1-m\right)=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow\left(m-1\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-4\left(tm\right)\end{matrix}\right.\)
Vậy m=-4
Câu 1:
\(1,\Leftrightarrow2x-2=3\Leftrightarrow x=\dfrac{5}{2}\\ 2,ĐK:x\ne\pm1\\ PT\Leftrightarrow\dfrac{2x^2+2x-1}{x^2-1}=2\\ \Leftrightarrow2x^2+2x-1=2x^2-2\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\\ 3,\Leftrightarrow\left[{}\begin{matrix}3x-2=2x-1\\3x-2=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)
\(4,\Leftrightarrow\left[{}\begin{matrix}3x-1=2-x\left(x\ge\dfrac{1}{3}\right)\\3x-1=x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(tm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\\ 5,\Leftrightarrow4x^2-2x+10=9x^2-6x+1\left(x\le\dfrac{1}{3}\right)\\ \Leftrightarrow5x^2-4x-9=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(6,\Leftrightarrow3x^2-9x+1=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow2x^2-5x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ 7,\Leftrightarrow2x^2+3x-4=7x+2\left(x\ge-\dfrac{2}{7}\right)\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)
Câu 2:
\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)
\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)
Câu 29: D
Câu 30: A
Câu 31: C