a.
Ta có: \(\sqrt{x^2+12}>\sqrt{x^2+5}\Rightarrow\sqrt{x^2+12}-\sqrt{x^2+5}>0\)
\(\Rightarrow3x-5=\sqrt{x^2+12}-\sqrt{x^2+5}>0\Rightarrow x>\dfrac{5}{3}\)
Do đó:
\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)
\(\Leftrightarrow3\left(x-2\right)+\sqrt{x^2+5}-3-\left(\sqrt{x^2+12}-4\right)=0\)
\(\Leftrightarrow3\left(x-2\right)+\dfrac{x^2-4}{\sqrt{x^2+5}+3}-\dfrac{x^2-4}{\sqrt{x^2+12}+4}=0\)
\(\Leftrightarrow\left(x-2\right)\left(3+\dfrac{x+2}{\sqrt{x^2+5}+3}-\dfrac{x+2}{\sqrt{x^2+12}+4}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[3+\left(x+2\right)\left(\dfrac{1}{\sqrt{x^2+5}+3}-\dfrac{1}{\sqrt{x^2+12}+4}\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(3+\dfrac{\left(x+2\right)\left(\sqrt{x^2+12}-\sqrt{x^2+5}+1\right)}{\left(\sqrt{x^2+5}+3\right)\left(\sqrt{x^2+12}+4\right)}\right)=0\)
\(\Leftrightarrow x-2=0\) (do \(x>\dfrac{5}{3}\) nên ngoặc phía sau luôn dương)
\(\Leftrightarrow x=2\)
b.
\(\Delta=\left(m-1\right)^2-4\left(-m^2+m-2\right)=5m^2-6m+9=5\left(m-\dfrac{3}{5}\right)^2+\dfrac{36}{5}>0\)
Phương trình đã cho luôn có 2 nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=-m^2+m-2\end{matrix}\right.\)
\(Q=\left(\dfrac{x_1}{x_2}\right)^2+\left(\dfrac{x_2}{x_1}\right)^2+2-2=\left(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}\right)^2-2\)
\(=\left(\dfrac{x_1^2+x_2^2}{x_1x_2}\right)^2-2=\left[\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}\right]^2-2\)
\(=\left[\dfrac{\left(m-1\right)^2-2\left(-m^2+m-2\right)}{-m^2+m-2}\right]^2-2=\left(\dfrac{3m^2-4m+5}{m^2-m+2}\right)^2-2\)
Ta có: \(\dfrac{3m^2-4m+5}{m^2-m+2}=\dfrac{3\left(m-\dfrac{2}{3}\right)^2+\dfrac{11}{3}}{\left(m-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}>0\)
\(\dfrac{3m^2-4m+5}{m^2-m+2}=\dfrac{21m^2-28m+35}{7\left(m^2-m+2\right)}=\dfrac{22\left(m^2-m+2\right)-\left(m^2+6m+9\right)}{7\left(m^2-m+2\right)}=\dfrac{22}{7}-\dfrac{\left(m+3\right)^2}{7\left(m^2-m+2\right)}\le\dfrac{22}{7}\)
\(\Rightarrow0< \dfrac{3m^2-4m+5}{m^2-m+2}\le\dfrac{22}{7}\)
\(\Rightarrow Q\le\left(\dfrac{22}{7}\right)^2-2\)
Dấu "=" xảy ra khi \(m+3=0\Leftrightarrow m=-3\)
