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Sagot :
Answer:
[tex]x=0[/tex] and [tex]x=1[/tex]
Step-by-step explanation:
[tex]\frac{x}{x-2}+\frac{x-1}{x+1}=-1[/tex]
[tex]\frac{x(x+1)}{(x-2)(x+1)}+\frac{(x-2)(x-1)}{(x-2)(x+1)}=-1[/tex]
[tex]\frac{x(x+1)+(x-2)(x-1)}{(x-2)(x+1)}=-1[/tex]
[tex]x(x+1)+(x-2)(x-1)=-(x-2)(x+1)[/tex]
[tex]x^2+x+x^2-3x+2=-(x^2-x-2)[/tex]
[tex]2x^2-2x+2=-x^2+x+2[/tex]
[tex]3x^2-2x+2=x+2[/tex]
[tex]3x^2-3x=0[/tex]
[tex]3x(x-1)=0[/tex]
[tex]x=0[/tex] and [tex]x=1[/tex]
Answer:
x = 0 or x = 1
Step-by-step explanation:
(x)/(x - 2) + (x - 1)/(x + 1) = -1
The LCD is (x - 2)(x + 1).
Multiply both sides of the equation by the LCD.
(x - 2)(x + 1) × (x)/(x - 2) + (x - 2)(x + 1) × (x - 1)/(x + 1) = (x - 2)(x + 1) × (-1)
The factors in bold cancel out of each product.
x(x + 1) + (x - 2)(x - 1) = -(x - 2)(x + 1)
x² + x + x² - x - 2x + 2 = -(x² + x - 2x - 2)
2x² - 2x + 2 = -x² + x + 2
3x² - 3x = 0
3x(x - 1) = 0
x = 0 or x = 1
The restrictions on the domain are:
x - 2 = 0
x = 2
x + 1 = 0
x = -1
Since our solutions do not include the excluded values, x = 2 and x = -1, both solutions x = 0 and x = 1 are valid.
Answer: x = 0 or x = 1
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