§3. Hàm số bậc hai

BPT có tập nghiệm là R khi và chỉ khi:

\(\left\{{}\begin{matrix}a=1>0\\\Delta'=m^2-4m-12\le0\end{matrix}\right.\)

\(\Rightarrow-2\le m\le6\)

Có \(6-\left(-2\right)+1=9\) giá trị nguyên của m

\(ycđb\Leftrightarrow\left[{}\begin{matrix}\Delta\le0\Leftrightarrow\left(m-5\right)^2\le0\Leftrightarrow m=5\\1\le x1< x2\left(1\right)\\x1< x2\le-1\left(2\right)\end{matrix}\right.\)

\(\Delta>0\Leftrightarrow\left(m-5\right)^2>0\Leftrightarrow m\ne5\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x1x2-\left(x1+x2\right)+1\ge0\\5m-5-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6m^2-10m-\left(5m-5\right)+1\ge0\\m>\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\le\dfrac{1}{2}\\m\ge2\end{matrix}\right.\\m>\dfrac{7}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m\ge2\\m\ne5\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x1+1\right)\left(x2+1\right)\ge0\\x1+x2+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x1x2+x1+x2+1\ge0\\x1+x2+2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6m^2-10m+5m-5+1\ge0\\5m-5-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\le-\dfrac{1}{2}\\m\ge\dfrac{4}{3}\end{matrix}\right.\\m< \dfrac{7}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m\in[\dfrac{4}{3};\dfrac{7}{5})\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\)\(m\in(-\infty;-\dfrac{1}{2}]\cup[\dfrac{4}{3};\dfrac{7}{5})\cup[2;+\infty)\cup\left\{5\right\}\)

Pt có 2 nghiệm khi: \(\left\{{}\begin{matrix}m\ne0\\\Delta'=9\left(m-1\right)^2-9m\left(m-3\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ge-1\end{matrix}\right.\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{6\left(m-1\right)}{m}\\x_1x_2=\dfrac{9\left(m-3\right)}{m}\end{matrix}\right.\)

\(x_1+x_2=x_1x_2\Rightarrow\dfrac{6\left(m-1\right)}{m}=\dfrac{9\left(m-3\right)}{m}\)

\(\Rightarrow6\left(m-1\right)=9\left(m-3\right)\)

\(\Rightarrow m=7\)

A đúng

b: \(\Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=2x\left(x+5\right)\)

\(\Leftrightarrow2x^2+10x+5x+25-2x^2-2x^2-10x=0\)

\(\Leftrightarrow-2x^2+5x+25=0\)

\(\Leftrightarrow2x^2-5x-25=0\)

\(\Rightarrow2x^2-10x+5x-25=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-5\right)=0\)

hay \(x\in\left\{-\dfrac{5}{2};5\right\}\)

a, ĐKXĐ:\(x\ge1\)

\(\sqrt{x-1}+3=x\\ \Leftrightarrow\sqrt{x-1}=x-3\left(x\ge3\right)\\ \Leftrightarrow x-1=x^2-6x+9\\ \Leftrightarrow x^2-7x+10=0\\ \Leftrightarrow\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=1\\ \Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=1\\ \Leftrightarrow2x^2+5x+10x+25-2x^2=2x^2+10x\\ \Leftrightarrow2x^2-5x-25=0\\ \Leftrightarrow\left(2x+5\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Chọn A

\(\left(-3;-1\right)\)

Chọn B

Chọn D