a) \(x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(\text{𝑥 ( 𝑥 − 1 ) − 3 𝑥 + 3 = 0}\)
⇔\(x^2-x-3x+3=0\)
⇔\(x^2-4x+4-1=0\)
⇔\(\left(x-2\right)^2-1^2=0\)
⇔\(\left(x-3\right).\left(x-1\right)\)=0
⇔\(\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x=3;x=1\)
#Fiona
Chúc bạn học tốt !
𝑥(𝑥−2)−1(𝑥−3)2=−1
⇔ \(x^2\)−2𝑥−1\(\left(x-3\right)^2\)= −1
⇔ \(x^2\)−2𝑥−1(𝑥−3)(𝑥−3)= −1
⇔ \(x^2\)−2𝑥−1[𝑥(𝑥−3)−3(𝑥−3)]= −1
⇔ \(x^2\)−2𝑥−1[\(x^2\)−3𝑥−3(𝑥−3)]= −1
⇔ \(x^2\)−2𝑥−1(\(x^2\)−3𝑥−3𝑥+9)= −1
⇔ \(x^2\)−2𝑥−1(\(x^2\)−6𝑥+9)= −1
⇔ \(x^2\)−2𝑥−\(x^2\)+6𝑥−9= −1
⇔ −2𝑥+6𝑥−9= −1
⇔ 4𝑥−9=−1
⇔ 4\(x\) = -1 + 9
⇔ \(x\) = 8 : 4
⇔ \(x\) = 2
Vậy \(x\)=2
#Fiona
Chúc bạn học tốt !