pt hoành độ giao điểm của \(\left(P\right):y=x^2\) và \(\left(d\right):y=2x+3\) là \(x^2=2x+3\Leftrightarrow x^2-2x-3=0\) \(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\).

 Khi \(x=3\) thì \(y=x^2=9\), khi \(x=-1\) thì \(y=x^2=1\). Do đó (P) cắt (d) tại \(A\left(3;9\right)\) và \(B\left(-1;1\right)\). Từ đó dễ dàng suy ra \(C\left(3;0\right)\) và \(D\left(-1;0\right)\). Từ đó suy ra \(CD=4\).

  Lại có \(AC=1;BD=9\). Do đó \(S_{ABCD}=\dfrac{\left(AC+BD\right).CD}{2}=\dfrac{\left(1+9\right).4}{2}=20\) (đơn vị diện tích)

a:

b:

\(\left|4\cdot\overrightarrow{a}-3\cdot\overrightarrow{b}\right|=\sqrt{13}\)

=>\((4\vec{a} - 3\vec{b})^2 = 13\)

=>\(16\vert{}\vec{a}\vert{}^2-24(\vec{a}\cdot\vec{b})+9\vert{}\vec{b}\vert{}^2=13\)

=>\(16\cdot1^2-24(\vec{a}\cdot\vec{b})+9\cdot^2=13\)

=>\(25 - 24(\vec{a} \cdot \vec{b}) = 13\)

=>\(24\cdot\overrightarrow{a}\cdot\overrightarrow{b}=12\)

=>\(\overrightarrow{a}\cdot\overrightarrow{b}=\frac12\)

\(\overrightarrow{c}\) ⊥\(\left(\overrightarrow{a}-\overrightarrow{b}\right)\)

=>\(\vec{c} \cdot (\vec{a} - \vec{b}) = 0\)

=>\((x\vec{a}+y\vec{b})\cdot(\vec{a}-\vec{b})=0\)

=>\(x\vert{}\vec{a}\vert{}^2-x(\vec{a}\cdot\vec{b})+y(\vec{a}\cdot\vec{b})-y\vert{}\vec{b}\vert{}^2=0\)

=>\(x - \frac{1}{2}x + \frac{1}{2}y - y = 0\)

=>x=y

\(\vec{c} = x\vec{a} + x\vec{b} = x(\vec{a} + \vec{b})\)

\(\vert{}\vec{c}\vert{}^2 = x^2(\vec{a} + \vec{b})^2 = 1\)

=>\(x^2(\vert{}\vec{a}\vert{}^2+2\vec{a}\cdot\vec{b}+\vert{}\vec{b}\vert{}^2)=1\)

=>\(x^2 \left(1 + 2 \cdot \frac{1}{2} + 1\right) = 1\)

=>\(x^2=\frac13\)

=>\(x=\pm\frac{\sqrt3}{3}\)

Khi \(x=\frac{\sqrt3}{3}\) thì \(y=x=\frac{\sqrt3}{3}\)

Khi \(x=-\frac{\sqrt3}{3}\) thì \(y=x=-\frac{\sqrt3}{3}\)