Giải thích các bước giải:
a,a,(2x+1) ²-2x-1=2
⇔ (2x+1).(2x+1)-(2x+1)=2
⇔ (2x+1)(2x+1-1)=2
⇔ 2x(2x+1)=2
⇔ x(2x+1)=0
⇔ \(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\2x=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=\frac{-1}{2} \end{array} \right.\)
Vậy x∈{0;$\frac{-1}{2}$}
b, (x²-3x)²+5(x²-3x)+6=0
⇔[(x²-3x)²+5(x²-3x)+$\frac{25}{4}$]²-$\frac{1}{4}$=0
⇔ [(x²-3x)+$\frac{5}{2}$]²-$\frac{1}{4}$=0
⇔ (x²-3x+$\frac{5}{2}$-$\frac{1}{2}$)(x²-3x+$\frac{5}{2}$+$\frac{1}{2}$)=0
⇔ \(\left[ \begin{array}{l}x²-3x+2=0\\x²-3x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}(x-2)(x-1)=0\\(x²-3x+\frac{9}{4})+\frac{3}{4}=0 (ktm)\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy x∈{2;1}
c, (x ² -x-1)(x ²-x)-2=0
⇔ x^4-x³-x³+x²-x²+x-2=0
⇔ x^4-2x³+x-2
⇔ x³(x-2)+(x-2)=0
⇔ (x³+1)(x-2)=0
⇔\(\left[ \begin{array}{l}x³+1=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy x∈{2;-1}
