Question

[Ly Po]

Giải pt bằng cách đặt 1 ẩn phụ

x²+\(\sqrt{x+1}=1\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge-1\)

\(x^2-1+\sqrt{x+1}=0\Rightarrow\left(x-1\right)\left(x+1\right)+\sqrt{x+1}=0\)

\(\Rightarrow\left(x+1-2\right)\left(x+1\right)+\sqrt{x+1}=0\)

Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x+1=t^2\) ta được:

\(\left(t^2-2\right)t^2+t=0\Rightarrow t\left(\left(t^2-2\right)t+1\right)=0\)

\(\Rightarrow t\left(t^3-2t+1\right)=0\Rightarrow t\left(t-1\right)\left(t^2+t-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=0\\t-1=0\\t^2+t-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}t=0\\t=1\\t=\dfrac{-1+\sqrt{5}}{2}\\t=\dfrac{-1-\sqrt{5}}{2}< 0\left(l\right)\end{matrix}\right.\)

TH1: \(t=0\Rightarrow\sqrt{x+1}=0\Rightarrow x=-1\)

TH2: \(t=1\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)

TH3: \(t=\dfrac{-1+\sqrt{5}}{2}\Rightarrow\sqrt{x+1}=\dfrac{-1+\sqrt{5}}{2}\Rightarrow x+1=\dfrac{3-\sqrt{5}}{2}\)

\(\Rightarrow x=\dfrac{3-\sqrt{5}}{2}-1=\dfrac{1-\sqrt{5}}{2}\)

Vậy pt có 3 nghiệm \(\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

Answer 2

[Akai Haruma]

Lời giải:

Đặt \(\sqrt{x+1}=a\Rightarrow 1=a^2-x\)

PT trở thành: \(x^2+a=a^2-x\)

\(\Leftrightarrow x^2-a^2+(a+x)=0\)

\(\Leftrightarrow (x+a)(x-a+1)=0\Rightarrow \left[\begin{matrix} x=-a\\ x+1=a\end{matrix}\right.\)

Nếu \(x=-a=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1+\sqrt{5}}{2}\)

Nếu \(x+1=a=\sqrt{x+1}\Rightarrow (x+1)^2=(x+1)\Rightarrow x(x+1)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\) (đều thỏa mãn)

Vậy.........