\(\Leftrightarrow x^2-4x+3=6\sqrt{2x+3}-18\)ĐK:\(x\ge\frac{-3}{2}\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-6\left(\frac{6-2x}{3+\sqrt{2x+3}}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x-1+\frac{12x-36}{3+\sqrt{2x+3}}\right]=0\)
Ta thấy bthức trong ngoặc vuông lớn hơn 0 với\(x\ge\frac{-3}{2}\)
Vậy x=3.
x2 - 4x + 21 = 6
<=> x2 - 4x + 21 - 6 = 0
<=> x2 - 4x + 15 = 0 (1)
\(\Delta\)' = (-2)2 - 1. 15 = -11 < 0
=> Pt (1) vô nghiệm
Vậy pt đã cho vô nghiệm
\(x\ge-\frac{3}{2}\)
\(x^2-6x+9+2x+3-6\sqrt{2x+3}+9=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{2x+3}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\) \(\Rightarrow x=3\)
