Câu 1:
Đặt \(x+1=a\). Khi đó \(x+3=a+2; x-1=a-2\).
PT đã cho tương đương với:
\((a+2)^4+(a-2)^4=626\)
\(\Leftrightarrow 2a^4+48a^2+32=626\)
\(\Leftrightarrow a^4+24a^2-297=0\)
\(\Leftrightarrow (a^2+12)^2=441\)
\(\Rightarrow a^2+12=\sqrt{441}=21\) (do \(a^2+12>0)\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Nếu $a=3$ thì \(x=a-1=2\)
Nếu $a=-3$ thì $x=a-1=-4$
Câu 2:
Đặt \(2x-1=a; x-1=b\). PT đã cho tương đương với:
\(a^3+b^3+(-a-b)^3=0\)
\(\Leftrightarrow a^3+b^3-(a+b)^3=0\)
\(\Leftrightarrow a^3+b^3-[a^3+b^3+3ab(a+b)]=0\)
\(\Leftrightarrow ab(a+b)=0\Rightarrow \left[\begin{matrix} a=0\\ b=0\\ a+b=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-1=0\\ 3x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\\ x=\frac{2}{3}\end{matrix}\right.\)
Câu 3:
PT \(\Leftrightarrow x^3(2x-1)-6x^2(2x-1)+9x(2x-1)-2(2x-1)=0\)
\(\Leftrightarrow (2x-1)(x^3-6x^2+9x-2)=0\)
\(\Leftrightarrow (2x-1)[x^2(x-2)-4x(x-2)+(x-2)]=0\)
\(\Leftrightarrow (2x-1)(x-2)(x^2-4x+1)=0\)
\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-2=0\\ x^2-4x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=2\\ (x-2)^2=3\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=2\\ x=2\pm \sqrt{3}\end{matrix}\right.\)
