Question

Giải phương trình :

\(x^3+\left(x+1\right)\sqrt{x+1}+2\sqrt{2}=\left(x+\sqrt{x+1}+\sqrt{2}\right)^3\)

Answer 1

ĐK:x\(\ge-1\)

\(x^3+\left(x+1\right)\sqrt{x+1}+2\sqrt{2}=\left(x+\sqrt{x+1}+\sqrt{2}\right)^3\)(*)

Đặt \(a=x;b=\sqrt{x+1};c=\sqrt{2}\)

Vậy (*)\(\Leftrightarrow a^3+b^3+c^3=\left(a+b+c\right)^3\Leftrightarrow3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

_a=-b\(\Leftrightarrow x=-\sqrt{x+1}\)\(\left(x\le0\right)\Leftrightarrow x^2=x+1\Leftrightarrow x^2-x-1=0\Leftrightarrow x=\dfrac{1-\sqrt{5}}{2}\)

_b=-c\(\Leftrightarrow\sqrt{x+1}=-\sqrt{2}\)(vô nghiệm)

_c=-a\(\Leftrightarrow\sqrt{2}=-x\Leftrightarrow x=-\sqrt{2}\left(ktm\right)\)

Vậy S={\(\dfrac{1-\sqrt{5}}{2}\)}