a) |x-2|+3x-9=0 b) (x^2-5x+1)^2 – 2x^2 + 10x = 1 Mn giúp mình nhé 01/10/2021 Bởi Quinn a) |x-2|+3x-9=0 b) (x^2-5x+1)^2 – 2x^2 + 10x = 1 Mn giúp mình nhé
a, $|x-2|+3x-9=0$ (ĐK: $x ≤ 3$) $↔|x-2|=-3x+9$ $↔$\(\left[ \begin{array}{l}x-2=-3x+9\\x-2=3x-9\end{array} \right.\) $↔$\(\left[ \begin{array}{l}x=\dfrac{11}{4}(\text{Nhận})\\x=\dfrac{7}{2}(\text{Loại})\end{array} \right.\) Vậy: `S={\frac{11}{4}}` b, $(x^2-5x+1)^2-2x^2+10x=1$ $↔x^4-10x^3+27x^2-10x+1-2x^2+10x-1=0$ $↔x^4-10x^3+25x^2=0$ $↔x^2(x-5)=0$ $↔$\(\left[ \begin{array}{l}x^2=0\\x-5=0\end{array} \right.\) $↔$\(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\) Vậy: `S={0;5}` Bình luận
a, $|x-2|+3x-9=0$ (ĐK: $x ≤ 3$)
$↔|x-2|=-3x+9$
$↔$\(\left[ \begin{array}{l}x-2=-3x+9\\x-2=3x-9\end{array} \right.\)
$↔$\(\left[ \begin{array}{l}x=\dfrac{11}{4}(\text{Nhận})\\x=\dfrac{7}{2}(\text{Loại})\end{array} \right.\)
Vậy: `S={\frac{11}{4}}`
b, $(x^2-5x+1)^2-2x^2+10x=1$
$↔x^4-10x^3+27x^2-10x+1-2x^2+10x-1=0$
$↔x^4-10x^3+25x^2=0$
$↔x^2(x-5)=0$
$↔$\(\left[ \begin{array}{l}x^2=0\\x-5=0\end{array} \right.\)
$↔$\(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\)
Vậy: `S={0;5}`
Đáp án:
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