Chương 2: HÀM SỐ BẬC NHẤT VÀ BẬC HAI

Chọn C

\(2,\Leftrightarrow\left|2x-4\right|=4-2x=-\left(2x-4\right)\\ \Leftrightarrow2x-4\le0\\ \Leftrightarrow x\le2\\ 3,\Leftrightarrow\left|\left(x+4\right)\left(x+1\right)\right|=x+4\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+4\right)\left(x+1\right)=x+4\left(x\le-4;x\ge-1\right)\\\left(x+4\right)\left(x+1\right)=-\left(x+4\right)\left(-4< x< -1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x+4\right)=0\left(x\le-4;x\ge-1\right)\\\left(x+4\right)\left(x+2\right)=0\left(-4< x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x^2-3x-4\\-2x-1=x^2-3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-5x-5=0\\x^2-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+3\sqrt{5}}{2}\\x=\dfrac{5-3\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{13}}{2}\\x=\dfrac{1-\sqrt{13}}{2}\end{matrix}\right.\)

\(5,Đk:x\ge3\\ PT\Leftrightarrow2x-3=x^2-6x+9\\ \Leftrightarrow x^2-8x+12=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=6\\ 6,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-\left(x-2\right)=0\\ \Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=2\)

\(7,ĐK:\left\{{}\begin{matrix}1-2x\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\in\varnothing\\ 8,ĐK:x\ge-\dfrac{1}{3}\\ PT\Leftrightarrow12x+4=x^2+4x+4\\ \Leftrightarrow x^2-8x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

D=R\{1}

D=R

Đặt \(\sqrt{x^2+4x+5}=t\Rightarrow t\in\left[\sqrt{5};\sqrt{17}\right]\)

\(\Rightarrow y=f\left(t\right)=t^2-2t+7\)

\(-\dfrac{b}{2a}=1\notin\left[\sqrt{5};\sqrt{17}\right]\)

\(f\left(\sqrt{5}\right)=10+4\sqrt{5}\) ; \(f\left(\sqrt{17}\right)=22+4\sqrt{17}\)

\(\Rightarrow y_{min}=10+4\sqrt{5}\) ; \(y_{max}=22+4\sqrt{17}\)

\(\left\{{}\begin{matrix}c=3\\-\dfrac{b}{2a}=1\\\dfrac{4ac-b^2}{4a}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}c=3\\b=-2a\\4ac-b^2=-8a\end{matrix}\right.\)

\(\Rightarrow12a-\left(-2a\right)^2=-8a\Rightarrow a=5\) \(\Rightarrow b=-10\)

\(\Rightarrow f\left(x\right)=5x^2-10x+3\)

Đồ thị hàm \(y=\left|5x^2-10x+3\right|\)

Từ đồ thị ta thấy \(y=m\) cắt \(y=\left|f\left(x\right)\right|\) tại 4 điểm pb hay pt đã cho có 4 nghiệm pb khi và chỉ khi: \(0< m< 2\)