Question

Giái phương trình:

a, \(\left(x^2+x+1\right)\left(6-2x\right)=0\)

b,\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)

c,\(\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

Answer 1

a, (x² + x + 1)(6 - 2x) = 0

Ta có: x² + x + 1 \(\Leftrightarrow\) (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)

Vì (x + \(\frac{1}{2}\))² \(\ge\) 0 với mọi x nên (x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) \(\ge\) \(\frac{3}{4}\) > 0 với mọi x

\(\Rightarrow\) 6 - 2x = 0

\(\Leftrightarrow\) x = 3

Vậy S = {3}

b, (x - 1)² - 1 + x² = (1 - x)(x + 3)

\(\Leftrightarrow\) (x - 1)² + (x² - 1) - (1 - x)(x + 3) = 0

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

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

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

\(\Leftrightarrow\) x - 1 = 0 hoặc 3x + 3 = 0

\(\Leftrightarrow\) x = 1 và x = -1

Vậy S = {1; -1}

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

\(\Leftrightarrow\) (x² - 1)(x + 2)(x - 3) - (x - 1)(x² - 4)(x + 5) = 0

\(\Leftrightarrow\) (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 - 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!!