giải các pt sau `(x^2+2x)^2-x^2-2x-2=0` `x^4 – x^3 – x^2 – x – 2 = 0 ` `x^3 – 2x^2 – 9x + 18 = 0` 12/08/2021 Bởi Camila giải các pt sau `(x^2+2x)^2-x^2-2x-2=0` `x^4 – x^3 – x^2 – x – 2 = 0 ` `x^3 – 2x^2 – 9x + 18 = 0`
$(x^2+2x)^2-x^2-2x-2=0$ $(1)$ $(x^2+2x)^2-(x^2+2x+2)=0$ Đặt $x^2+2x=t$ Từ $(1)⇒t^2-t-2=0$ $⇔(t^2-2t)+(t-2)=0$ $⇔(t-2)(t+1)=0$ $⇔\left[ \begin{array}{l}t-2=0\\t+1=0\end{array} \right.⇔\left[ \begin{array}{l}t=2\\t=-1\end{array} \right.$ Với $t=2⇒x^2+2x-2=0$ $⇔(x^2+2x+1)-3=0$ $⇔(x+1)^2-3=0$$⇔(x+1+\sqrt{3})(x+1-\sqrt{3})=0$ $⇔\left[ \begin{array}{l}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{array} \right.$ Với $t=-1⇒x^2+2x+1=0$ $⇔(x+1)^2=0$ $⇔x=-1$ Vậy $S=\{-1-\sqrt{3};-1+\sqrt{3};-1\}$ $x^4-x^3-x^2-x-2=0$ $⇔(x^4-2x^3)+(x^3-2x^2)+(x^2-2x)+(x-2)=0$ $⇔(x-2)(x^3+x^2+x+1)=0$ $⇔(x-2)[(x^3+x^2)+(x+1)]=0$ $⇔(x-2)(x+1)(x^2+1)=0$ Vì $x^2≥0∀x⇒x^2+1>0∀x$ $⇒(x-2)(x+1)=0$ $⇔\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$ Vậy $S=\{2;-1\}$ $x^3-2x^2-9x+18=0$ $⇔(x^3-2x^2)-(9x-18)=0$ $⇔(x-2)(x^2-9)=0$ $⇔(x-2)(x+3)(x-3)=0$ $⇔\left[ \begin{array}{l}x-2=0\\x-3=0\\x+3=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=3\\x=-3\end{array} \right.$ Vậy $S=\{2;3;-3\}$. Bình luận
`(x^2+2x)^2−x^2−2x−2=0 (1)` `(x^2+2x)^2−(x^2+2x+2)=0` Đặt `x^2+2x=t` Từ `(1)⇒t^2−t−2=0` ⇔`(t^2−2t)+(t−2)=0` ⇔`(t−2)(t+1)=0` ⇔`t−2=0 hoặc t+1=0` ⇔`t=2 hoặc t=−1` Với `t=2⇒x^2+2x−2=0` ⇔`(x^2+2x+1)−3=0` ⇔`(x+1)^2−3=0`⇔`(x+1+√3)(x+1−√3)=0` ⇔`x=−1−√3` hoặc `x=−1+√3` Với `t=−1⇒x^2+2x+1=0` ⇔`(x+1)^2=0` ⇔`x=−1` Vậy `S={−1−√3;−1+√3;−1}` `x^4−x^3−x^2−x−2=0` ⇔`(x^4−2x^3)+(x^3−2x^2)+(x^2−2x)+(x−2)=0` ⇔`(x−2)(x^3+x^2+x+1)=0` ⇔`(x−2)[(x^3+x^2)+(x+1)]=0` ⇔`(x−2)(x+1)(x2+1)=0` Vì `x^2≥0∀x` ⇒`x^2+1>0∀x` ⇒`(x−2)(x+1)=0` ⇔`x−2=0` hoặc `x+1=0` ⇔`x=2` hoặc `x=−1` Vậy `S={2;−1}` `x^3−2x^2−9x+18=0` ⇔`(x^3−2x^2)−(9x−18)=0` ⇔`(x−2)(x^2−9)=0` ⇔`(x−2)(x+3)(x−3)=0` ⇔`x−2=0` hoặc `x−3=0` hoặc `x+3=0` ⇔`x=2` hoặc `x=3` hoặc `x=−3` Vậy `S={2;3;−3}` Bình luận
$(x^2+2x)^2-x^2-2x-2=0$ $(1)$
$(x^2+2x)^2-(x^2+2x+2)=0$
Đặt $x^2+2x=t$
Từ $(1)⇒t^2-t-2=0$
$⇔(t^2-2t)+(t-2)=0$
$⇔(t-2)(t+1)=0$
$⇔\left[ \begin{array}{l}t-2=0\\t+1=0\end{array} \right.⇔\left[ \begin{array}{l}t=2\\t=-1\end{array} \right.$
Với $t=2⇒x^2+2x-2=0$
$⇔(x^2+2x+1)-3=0$
$⇔(x+1)^2-3=0$
$⇔(x+1+\sqrt{3})(x+1-\sqrt{3})=0$
$⇔\left[ \begin{array}{l}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{array} \right.$
Với $t=-1⇒x^2+2x+1=0$
$⇔(x+1)^2=0$
$⇔x=-1$
Vậy $S=\{-1-\sqrt{3};-1+\sqrt{3};-1\}$
$x^4-x^3-x^2-x-2=0$
$⇔(x^4-2x^3)+(x^3-2x^2)+(x^2-2x)+(x-2)=0$
$⇔(x-2)(x^3+x^2+x+1)=0$
$⇔(x-2)[(x^3+x^2)+(x+1)]=0$
$⇔(x-2)(x+1)(x^2+1)=0$
Vì $x^2≥0∀x⇒x^2+1>0∀x$
$⇒(x-2)(x+1)=0$
$⇔\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$
Vậy $S=\{2;-1\}$
$x^3-2x^2-9x+18=0$
$⇔(x^3-2x^2)-(9x-18)=0$
$⇔(x-2)(x^2-9)=0$
$⇔(x-2)(x+3)(x-3)=0$
$⇔\left[ \begin{array}{l}x-2=0\\x-3=0\\x+3=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=3\\x=-3\end{array} \right.$
Vậy $S=\{2;3;-3\}$.
`(x^2+2x)^2−x^2−2x−2=0 (1)`
`(x^2+2x)^2−(x^2+2x+2)=0`
Đặt `x^2+2x=t`
Từ `(1)⇒t^2−t−2=0`
⇔`(t^2−2t)+(t−2)=0`
⇔`(t−2)(t+1)=0`
⇔`t−2=0 hoặc t+1=0`
⇔`t=2 hoặc t=−1`
Với `t=2⇒x^2+2x−2=0`
⇔`(x^2+2x+1)−3=0`
⇔`(x+1)^2−3=0`
⇔`(x+1+√3)(x+1−√3)=0`
⇔`x=−1−√3` hoặc `x=−1+√3`
Với `t=−1⇒x^2+2x+1=0`
⇔`(x+1)^2=0`
⇔`x=−1`
Vậy `S={−1−√3;−1+√3;−1}`
`x^4−x^3−x^2−x−2=0`
⇔`(x^4−2x^3)+(x^3−2x^2)+(x^2−2x)+(x−2)=0`
⇔`(x−2)(x^3+x^2+x+1)=0`
⇔`(x−2)[(x^3+x^2)+(x+1)]=0`
⇔`(x−2)(x+1)(x2+1)=0`
Vì `x^2≥0∀x`
⇒`x^2+1>0∀x`
⇒`(x−2)(x+1)=0`
⇔`x−2=0` hoặc `x+1=0`
⇔`x=2` hoặc `x=−1`
Vậy `S={2;−1}`
`x^3−2x^2−9x+18=0`
⇔`(x^3−2x^2)−(9x−18)=0`
⇔`(x−2)(x^2−9)=0`
⇔`(x−2)(x+3)(x−3)=0`
⇔`x−2=0` hoặc `x−3=0` hoặc `x+3=0`
⇔`x=2` hoặc `x=3` hoặc `x=−3`
Vậy `S={2;3;−3}`