Question

Answer 1

a) \(x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Answer 2

\(\text{𝑥 ( 𝑥 − 1 ) − 3 𝑥 + 3 = 0}\)

⇔\(x^2-x-3x+3=0\)

⇔\(x^2-4x+4-1=0\)

⇔\(\left(x-2\right)^2-1^2=0\)

⇔\(\left(x-3\right).\left(x-1\right)\)=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

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Answer 3

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

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