Question

a) x(x+1) –5x–5=0

b)2x²–5x–12=0

Answer 1

a) x(x + 1) – 5x – 5 = 0

<=> x(x + 1) - 5(x + 1) = 0

<=> (x + 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)

Vậy \(S=\left\{-1;5\right\}\)

b) 2x2 – 5x – 12 = 0

<=> 2x² – 8x + 3x – 12 = 0

<=> (2x² - 8x) + (3x - 12) = 0

<=> 2x(x - 4) + 3(x - 4) = 0

<=> (x - 4)(2x + 3) = 0

<=> \(\left[{}\begin{matrix}x-4=0\\2x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=4\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(S=\left\{4;-\frac{3}{2}\right\}\)

#Học tốt!

Answer 2

a, x(x + 1) - 5x - 5 = 0

\(\Leftrightarrow\) x(x + 1) - 5(x + 1) = 0

\(\Leftrightarrow\) (x + 1)(x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)

Vậy S = {-1; 5}

phần b để mình nghĩ tiếp đã

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