(x+3)^3-3x^2-x^3+9=x-8 (x+3)(x-3)+8x-2=2x+1 14/09/2021 Bởi Bella (x+3)^3-3x^2-x^3+9=x-8 (x+3)(x-3)+8x-2=2x+1
Đáp án: $\text{a) (x+3)³ – 3x² -x³ +9 = x – 8 }$ $\text{⇔ x³ + 27x + 9x² +27 -3x² -x³ +9 = x – 8 }$ $\text{⇔ x³ -x³ +9x² -3x² +27x -x +27 +9 +8=0 }$ $\text{⇔ 6x² +26x +44 =0 }$ $\text{⇔ (√6x)² + 2.√6x.$\dfrac{13\sqrt[]{6}}{6}$ + ($\dfrac{13\sqrt[]{6}}{6}$)² – ($\dfrac{13\sqrt[]{6}}{6}$)² +44 =0 }$ $\text{⇔ (√6x + $\dfrac{13\sqrt[]{6}}{6}$)² + $\dfrac{95}{6}$ =0 }$ $\text{Mà (√6x + $\dfrac{13\sqrt[]{6}}{6}$) ² ≥ 0 }$ $\text{⇒ (√6x + $\dfrac{13\sqrt[]{6}}{6}$)² + $\dfrac{95}{6}$ >0 ∀ x ∈ R }$ $\text{⇒ x vô nghiệm }$ $\text{Vậy x ∈ { ∅} }$ $\text{b)(x+3)(x-3) + 8x-2 =2x +1 }$ $\text{⇔ x² -9 + 8x -2 =2x +1 }$ $\text{⇔ x² +8x -2x -9-2-1 =0 }$ $\text{⇔ x² +6x -12 =0}$ $\text{⇔ x² +2.x.3+(3)² -(3)² -12 =0 }$ $\text{⇔ (x +3)² – 21 =0 }$ $\text{⇔ (x+3 – √21) (x+3+√21) =0 }$ $\text{⇔ \(\left[ \begin{array}{l}x+3 – √21=0\\x+3+√21=0\end{array} \right.\) }$ $\text{⇔\(\left[ \begin{array}{l}x=-3+ \sqrt21\\x=-3-\sqrt21\end{array} \right.\) }$ $\text{Vậy x ∈ { -3+ $\sqrt21$ ; -3 – $\sqrt{21}$ } }$ Bình luận
`1) (x+3)^3-3x^2-x^3+9=x-8` `⇔(x^3+9x^2+27x+27)-3x^2-x^3+9=x-8` `⇔x^3+9x^2+27x+27-3x^2-x^3+9-x+8=0` `⇔6x^2+26x+44=0` `⇔36x^2+156x+264=0` `⇔(6x)^2+2.6x.13+169+95=0` `⇔(6x+13)^2=-95` ( vô lí, vì `(6x+13)^2\geq0>0> -95` ) Vậy không có x thỏa mãn. `2) (x+3)(x-3)+8x-2=2x+1` `⇔ x^2-9+8x-2-2x-1=0` `⇔ x^2+6x-12=0` `⇔ x^2+6x+9-21=0` $⇔ (x+3)^2- (\sqrt[]{21})^2=0$ $⇔ (x+3-\sqrt[]{21})(x+3+\sqrt[]{21})=0$ $⇒$ \(\left[ \begin{array}{l}x+3-\sqrt[]{21}=0\\x+3+\sqrt[]{21}=0\end{array} \right.\) $⇒$ \(\left[ \begin{array}{l}x=\sqrt[]{21}-3\\x=-\sqrt[]{21}-3\end{array} \right.\) Vậy x ∈ { $± \sqrt[]{21}-3$ }. Bình luận
Đáp án:
$\text{a) (x+3)³ – 3x² -x³ +9 = x – 8 }$
$\text{⇔ x³ + 27x + 9x² +27 -3x² -x³ +9 = x – 8 }$
$\text{⇔ x³ -x³ +9x² -3x² +27x -x +27 +9 +8=0 }$
$\text{⇔ 6x² +26x +44 =0 }$
$\text{⇔ (√6x)² + 2.√6x.$\dfrac{13\sqrt[]{6}}{6}$ + ($\dfrac{13\sqrt[]{6}}{6}$)² – ($\dfrac{13\sqrt[]{6}}{6}$)² +44 =0 }$
$\text{⇔ (√6x + $\dfrac{13\sqrt[]{6}}{6}$)² + $\dfrac{95}{6}$ =0 }$
$\text{Mà (√6x + $\dfrac{13\sqrt[]{6}}{6}$) ² ≥ 0 }$
$\text{⇒ (√6x + $\dfrac{13\sqrt[]{6}}{6}$)² + $\dfrac{95}{6}$ >0 ∀ x ∈ R }$
$\text{⇒ x vô nghiệm }$
$\text{Vậy x ∈ { ∅} }$
$\text{b)(x+3)(x-3) + 8x-2 =2x +1 }$
$\text{⇔ x² -9 + 8x -2 =2x +1 }$
$\text{⇔ x² +8x -2x -9-2-1 =0 }$
$\text{⇔ x² +6x -12 =0}$
$\text{⇔ x² +2.x.3+(3)² -(3)² -12 =0 }$
$\text{⇔ (x +3)² – 21 =0 }$
$\text{⇔ (x+3 – √21) (x+3+√21) =0 }$
$\text{⇔ \(\left[ \begin{array}{l}x+3 – √21=0\\x+3+√21=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=-3+ \sqrt21\\x=-3-\sqrt21\end{array} \right.\) }$
$\text{Vậy x ∈ { -3+ $\sqrt21$ ; -3 – $\sqrt{21}$ } }$
`1) (x+3)^3-3x^2-x^3+9=x-8`
`⇔(x^3+9x^2+27x+27)-3x^2-x^3+9=x-8`
`⇔x^3+9x^2+27x+27-3x^2-x^3+9-x+8=0`
`⇔6x^2+26x+44=0`
`⇔36x^2+156x+264=0`
`⇔(6x)^2+2.6x.13+169+95=0`
`⇔(6x+13)^2=-95` ( vô lí, vì `(6x+13)^2\geq0>0> -95` )
Vậy không có x thỏa mãn.
`2) (x+3)(x-3)+8x-2=2x+1`
`⇔ x^2-9+8x-2-2x-1=0`
`⇔ x^2+6x-12=0`
`⇔ x^2+6x+9-21=0`
$⇔ (x+3)^2- (\sqrt[]{21})^2=0$
$⇔ (x+3-\sqrt[]{21})(x+3+\sqrt[]{21})=0$
$⇒$ \(\left[ \begin{array}{l}x+3-\sqrt[]{21}=0\\x+3+\sqrt[]{21}=0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=\sqrt[]{21}-3\\x=-\sqrt[]{21}-3\end{array} \right.\)
Vậy x ∈ { $± \sqrt[]{21}-3$ }.