giúp vs
A3:
a. \(x^2-x+2< 0\)
Ta có: \(\left[{}\begin{matrix}\Delta=\left(-1\right)^2-4\cdot1\cdot2=-7< 0\\f\left(x\right)=0\Leftrightarrow x^2-x+2=0\left(vo\cdot nghiem\right)\end{matrix}\right.\)
Vì \(a>0\Rightarrow f\left(x\right)>0\forall x\in R\)
b. \(4x^2-4x+1>0\)
Ta có: \(\left[{}\begin{matrix}\Delta=\left(-4\right)^2-4\cdot4\cdot1=0=0\\f\left(x\right)=0\Leftrightarrow....\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)
Vì \(a>0\Rightarrow f\left(x\right)>0\forall x\ne\dfrac{1}{2}\)
B4:
a. Ta có: \(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(1-0;0-3\right)=\left(1;-3\right)\\\overrightarrow{AC}=\left(-3-0;2-3\right)=\left(-3;-1\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}AB=\sqrt{\left(1\right)^2+\left(-3\right)^2}=\sqrt{10}\\AC=\sqrt{\left(-3\right)^2+\left(-1\right)^2}=\sqrt{10}\end{matrix}\right.\)
Ta có: \(1\cdot\left(-1\right)-\left(-3\right)\cdot\left(-3\right)=-10\ne0\)
=> \(\overrightarrow{AB},\overrightarrow{AC}\) không cùng phương
=> A, B, C không thẳng hàng
=> A, B, C là 3 đỉnh của 1 tam giác
Mà \(AB=AC=\sqrt{10}\Rightarrow\Delta ABC\) cân tại A
b. Toạ độ trung điểm H của BC là \(H\left(x_H;y_H\right)\) có: \(\left\{{}\begin{matrix}x_H=\dfrac{1+\left(-3\right)}{2}=-1\\y_H=\dfrac{0+2}{2}=1\end{matrix}\right.\)
\(\Rightarrow H\left(-1;1\right)\)
Ta có: \(\left\{{}\begin{matrix}\overrightarrow{AH}=\left(-1-0;1-3\right)=\left(-1;-2\right)\\\overrightarrow{BC}=\left(-3-1;2-0\right)=\left(-4;2\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}AH=\sqrt{\left(-1\right)^2+\left(-2\right)^2}=\sqrt{5}\\BC=\sqrt{\left(-4\right)^2+2^2}=2\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot\sqrt{5}\cdot2\sqrt{5}=5\)
