Đáp án:
Giải thích các bước giải:
1.x ²(x ²+1)+x ²(x+3)+3x+3=0
⇔$x^{4}$ +x² +x³ +3x² +3x +3=0
⇔($x^{4}$ +x³+x²) +(3x² +3x+3)=0
⇔x².(x² +x+1) +3.(x² +x+1) =0
⇔(x² +x +1).(x² +3) =0
xét :x² +x +1=0
⇔(x² + $\frac{1}{2}$ .x.2+$\frac{1}{4}$ )+$\frac{3}{4}$ =0
⇔(x +$\frac{1}{2}$ )² +$\frac{3}{4}$ =0
vì (x +$\frac{1}{2}$ )² ≥0 nên (x +$\frac{1}{2}$ )² +$\frac{3}{4}$ >0
+) x² +3 >0 vì x² ≥0
vậy phương trình vô nghiệm.
2.2x ³+3x ²+2x+3= 0
⇔ (2x³ +2x) +(3x² +3) =0
⇔2x.(x² +1) +3.(x² +1) =0
⇔(x² +1).(2x +3) =0
⇔2x +3 =0 (vì x² +1 >0)
⇔2x =-3
⇔x =-$\frac{3}{2}$
3. x(2x-7)-4x+14=0
⇔x.(2x -7) -(4x -14) =0
⇔x.(2x -7) -2.(2x -7)=0
⇔(2x -7).(x -2)=0
⇔\(\left[ \begin{array}{l}2x -7=0\\x -2 =0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x =7\\x =2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =\frac{7}{2}\\x =2\end{array} \right.\)
