Question

giải pt

\(\dfrac{1}{x^2-3x+2}-\dfrac{x-5}{2-x}=\dfrac{3}{10}\)

Answer 1

x^2-3x+2=(x-1)(x-2)

dk x≠1;2

1+(x-5)(x-1)=3/10(x^2-3x+2)

10+10x^2-60x+50=3x^2-9x+6

7x^2-54x-54=0

x=(27±3√123)/7

Answer 2

\(\dfrac{1}{x^2-3x+2}-\dfrac{x-5}{2-x}=\dfrac{3}{10}\)

⇔ \(\dfrac{1}{x^2-x-2x+2}+\dfrac{x-5}{x-2}=\dfrac{3}{10}\)

⇔ \(\dfrac{10}{10\left(x-1\right)\left(x-2\right)}+\dfrac{10\left(x-5\right)\left(x-1\right)}{10\left(x-1\right)\left(x-2\right)}=\dfrac{3\left(x^2-3x+2\right)}{10\left(x-1\right)\left(x-2\right)}\)( x # 1 ; x # 2)

⇔ 10 + 10( x² - 6x + 5)= 3(x² - 3x + 2)

⇔ 10 + 10x² - 60x + 50 = 3x² - 9x + 6

⇔ 7x² - 51x - 54 = 0

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