Đặt \(\sqrt[3]{x-2}\) = a , \(\sqrt{x+1}=b\)
ta có: \(\left\{{}\begin{matrix}a+b=3\left(1\right)\\a^3-b^2=-1\left(2\right)\end{matrix}\right.\)
thay (1)vào(2) ta có
a3-(3-a)2+1=0
⇔a3-a2+6a-8=0
⇔\(\left[{}\begin{matrix}a=\\a=\end{matrix}\right.\)
Dat \(\left\{{}\begin{matrix}\sqrt[3]{x-2}=a\\\sqrt{x+1}=b\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=3\\a^3-b^2=-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}b=3-a\left(1\right)\\a^3-b^2=-3\left(2\right)\end{matrix}\right.\)
Tu (1)(2) => a3-(3-a)2=-3
=> a3-(9-6a+a2)=-3
=> a3-a2+6a-9+3=0
=> a3-a2+6a-6=0
=> a2(a-1)+6(a-1)=0
=> (a2+6)(a-1)=0
Vi a2+6>0 => a-1=0 => a=1 => b=2
Voi a=1 thi \(\sqrt[3]{x-2}=1\)
=> x-2=1 => x=3
Voi b=2 thi \(\sqrt{x+1}=2\)
=> x+1=4 => x=3
