a. √(x2+12) + 5 = 3x + √(x2+5
Dễ thấy, nếu x<0
\(VP=3x+\sqrt{x^2+5}< \sqrt{x^2+12}< \sqrt{x^2+12}+5\)
=>Pt vô nghiệm.Vậy \(x\ge0\)
\(pt\Leftrightarrow\left(\sqrt{x^2+5}-3\right)-\left(\sqrt{x^2+12}-4\right)+3x-6=0\)
\(\Leftrightarrow\frac{x^2-4}{\sqrt{x^2+5}+3}-\frac{x^2-4}{\sqrt{x^2+12}+4}+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}+3=0\left(2\right)\end{array}\right.\)
Ta có:
\(\left(2\right)\Leftrightarrow\left(x+2\right)\left(\frac{1}{\sqrt{x^2+5}+3}-\frac{1}{\sqrt{x^2+12}+4}\right)+3=0\)
\(\Leftrightarrow\left(x+2\right)\cdot\frac{\sqrt{x^2+12}-\sqrt{x^2+5}+1}{\left(\sqrt{x^2+5}+3\right)\left(\sqrt{x^2+12}+4\right)}=0\)
Do x>0 nên VT>0=VP =>pt (2) vô nghiệm
Vậy phương trình ban đầu có nghiệm duy nhất x=2