1.x ²(x ²+1)+x ²(x+3)+3x+3=0 2.2x ³+3x ²+2x+3 =0 3. x(2x-7)-4x+14=0 03/08/2021 Bởi Elliana 1.x ²(x ²+1)+x ²(x+3)+3x+3=0 2.2x ³+3x ²+2x+3 =0 3. x(2x-7)-4x+14=0
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 ⇔\(\left[ \begin{array}{l}2x -7=0\\x -2 =0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x =7\\x =2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x =\frac{7}{2}\\x =2\end{array} \right.\) Bình luận
Đáp án: $1, x^2(x^2+1)+x^2(x+3)+3x+3=0$ $⇔ x^4 +x^2 +x^3 +3x^2 +3x+3=0$ $⇔x^4+x^3+4x^2 +3x+3=0$ $⇔(x^4+x^3+x^2)+(3x^2+3x+3)+0$ $⇔x^2(x^2+x+1)+3(x^2+x+1)=0$ $⇔(x^2+x+1)(x^2+3)=0$ Th1 : $x^2+x+1 =0$ $⇔x^2+2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} =0$ $⇔(x+\dfrac{1}{2})^2 + \dfrac{3}{4}$ =0$ Vì $(x+\dfrac{1}{2})^2 ≥ 0 ⇔(x+\dfrac{1}{2})^2 +\dfrac{3}{4} > 0$ (vô nghiệm) Th2 : $x^2+3=0$ Vì $x^2≥ 0 ⇔x^2+3 > 0 $ (vô nghiệm) $2 ) 2x^3 +3x^2+2x+3 =0$ $⇔ x^2(2x+3)+(2x+3)=0$ $⇔(2x+3)(x^2+1)=0$ Th1 : $x^2+1= 0$ Vì $x^2≥ 0 ⇔x^2+1 >0$ (vô nghiệm) Th2 : $2x+3=0$ $⇔2x=-3$ $⇔x = -\dfrac{3}{2}$ Vậy $x=-\dfrac{3}{2}$ $3) x(2x-7)-4x+14=0$ $⇔x(2x-7)-2(2x-7)=0$ $⇔(2x-7)(x-2)=0$ ⇔\(\left[ \begin{array}{l}2x-7=0\\x-2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=2\end{array} \right.\) Vậy $\text{x ∈ {$\dfrac{7}{2} ; 2$ }}$ Giải thích các bước giải: Bình luận
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
⇔\(\left[ \begin{array}{l}2x -7=0\\x -2 =0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x =7\\x =2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x =\frac{7}{2}\\x =2\end{array} \right.\)
Đáp án:
$1, x^2(x^2+1)+x^2(x+3)+3x+3=0$
$⇔ x^4 +x^2 +x^3 +3x^2 +3x+3=0$
$⇔x^4+x^3+4x^2 +3x+3=0$
$⇔(x^4+x^3+x^2)+(3x^2+3x+3)+0$
$⇔x^2(x^2+x+1)+3(x^2+x+1)=0$
$⇔(x^2+x+1)(x^2+3)=0$
Th1 : $x^2+x+1 =0$
$⇔x^2+2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} =0$
$⇔(x+\dfrac{1}{2})^2 + \dfrac{3}{4}$ =0$
Vì $(x+\dfrac{1}{2})^2 ≥ 0 ⇔(x+\dfrac{1}{2})^2 +\dfrac{3}{4} > 0$ (vô nghiệm)
Th2 : $x^2+3=0$
Vì $x^2≥ 0 ⇔x^2+3 > 0 $ (vô nghiệm)
$2 ) 2x^3 +3x^2+2x+3 =0$
$⇔ x^2(2x+3)+(2x+3)=0$
$⇔(2x+3)(x^2+1)=0$
Th1 : $x^2+1= 0$
Vì $x^2≥ 0 ⇔x^2+1 >0$ (vô nghiệm)
Th2 : $2x+3=0$
$⇔2x=-3$
$⇔x = -\dfrac{3}{2}$
Vậy $x=-\dfrac{3}{2}$
$3) x(2x-7)-4x+14=0$
$⇔x(2x-7)-2(2x-7)=0$
$⇔(2x-7)(x-2)=0$
⇔\(\left[ \begin{array}{l}2x-7=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=2\end{array} \right.\)
Vậy $\text{x ∈ {$\dfrac{7}{2} ; 2$ }}$
Giải thích các bước giải: