Question

(x^{^2}-1).(x+2).(x-3)=(x-1).(x^{^2}-4).(x+5)

Answer 1

(x² - 1)(x + 2)(x - 3) = (x - 1)(x² - 4)(x + 5)

\(\Leftrightarrow\) (x - 1)(x + 1)(x + 2)(x - 3) - (x - 1)(x - 2)(x + 2)(x + 5) = 0

\(\Leftrightarrow\) (x - 1)(x + 2)[(x + 1)(x - 3) - (x - 2)(x + 5)] = 0

\(\Leftrightarrow\) (x² + 2x - x - 2)(x² - 3x + x - 3 - x² - 5x + 2x + 10) = 0

\(\Leftrightarrow\) (x² + x - 2)(-5x + 7) = 0

\(\Leftrightarrow\) [(x + \(\frac{1}{2}\))² - \(\frac{9}{4}\)](-5x + 7) = 0

\(\Leftrightarrow\) (x + \(\frac{1}{2}\) - \(\frac{3}{2}\))(x + \(\frac{1}{2}\) + \(\frac{3}{2}\))(-5x + 7) = 0

\(\Leftrightarrow\) (x - 1)(x + 2)(-5x + 7) = 0

\(\Leftrightarrow\) x - 1 = 0 hoặc x + 2 = 0 hoặc -5x + 7 = 0

\(\Leftrightarrow\) x = 1 và x = -2 và x = \(\frac{7}{5}\)

Vậy S = {1; -2; \(\frac{7}{5}\)}

Chúc bn học tốt!!

Answer 2

Ta có: \(\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x^2-1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x^2-4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x^2-2x-3\right)-\left(x^2+3x-10\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(7-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\7-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\5x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{7}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-2;\frac{7}{5}\right\}\)