Lời giải:
Đặt \(\sqrt{x+2018}=a(a\geq 0)\Rightarrow 2018=a^2-x\)
PT đã cho trở thành:
\(x^2+a=a^2-x\)
\(\Leftrightarrow (x^2-a^2)+(a+x)=0\)
\(\Leftrightarrow (x+a)(x-a+1)=0\)
\(\Rightarrow \left[\begin{matrix} x+a=0\\ x-a+1=0\end{matrix}\right.\)
Nếu \(x+a=0\Rightarrow a=-x\Leftrightarrow \sqrt{x+2018}=-x\)
\(\Rightarrow \left\{\begin{matrix} x\leq 0\\ x+2018=x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\leq 0\\ x=\frac{1\pm 3\sqrt{897}}{2}\end{matrix}\right.\) (giải pt bậc 2 cơ bản)
\(\Rightarrow x=\frac{1-3\sqrt{897}}{2}\)
Nếu \(x-a+1=0\Rightarrow a=x+1\Rightarrow \sqrt{x+2018}=x+1\)
\(\Rightarrow \left\{\begin{matrix} x+2018=(x+1)^2\\ x\geq -1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x^2+x-2017=0\\ x\geq -1\end{matrix}\right.\)
\(\Rightarrow x=\frac{\sqrt{8069}-1}{2}\)
Lời giải
Đặt √x+2018=a(a≥0)⇒2018=a2−xx+2018=a(a≥0)⇒2018=a2−x
PT đã cho trở thành:
x2+a=a2−xx2+a=a2−x
⇔(x2−a2)+(a+x)=0⇔(x2−a2)+(a+x)=0
⇔(x+a)(x−a+1)=0⇔(x+a)(x−a+1)=0
⇒[x+a=0x−a+1=0⇒[x+a=0x−a+1=0
Nếu x+a=0⇒a=−x⇔√x+2018=−xx+a=0⇒a=−x⇔x+2018=−x
⇒{x≤0x+2018=x2⇒{x≤0x+2018=x2
⇒{x≤0x=1±3√8972⇒{x≤0x=1±38972 (giải pt bậc 2 cơ bản)
⇒x=1−3√8972⇒x=1−38972
Nếu x−a+1=0⇒a=x+1⇒√x+2018=x+1x−a+1=0⇒a=x+1⇒x+2018=x+1
⇒{x+2018=(x+1)2x≥−1⇒{x2+x−2017=0x≥−1⇒{x+2018=(x+1)2x≥−1⇒{x2+x−2017=0x≥−1
⇒x=√8069−12
\(x^2+\sqrt{x+2018}=2018\)
\(\Leftrightarrow\left(x^2+x-2017\right)+\sqrt{x+2018}-\left(x+1\right)=0\)
\(\Leftrightarrow x^2+x-2017-\dfrac{x^2+x-2017}{\sqrt{x+2018}+x+1}=0\)
\(\Leftrightarrow\left(x^2+x-2017\right)\left(1-\dfrac{1}{\sqrt{x+2018}+x+1}\right)=0\)
