Answer:
[tex]x = 2, 4[/tex]
Step-by-step explanation:
Given :
[tex]\frac{4-x}{x^{2} -5x+4} + \frac{2}{x-1} =1[/tex]
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Factorize the denominator of the first term :
⇒ x² - 5x + 4
⇒ x² - x - 4x + 4
⇒ x(x - 1) - 4(x - 1)
⇒ (x - 4)(x - 1)
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Hence, the equation now is :
[tex]\frac{4-x}{(x-1)(x-4)} + \frac{2}{x-1} =1[/tex]
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Multiply the second term by (x - 4) in the numerator and denominator :
[tex]\frac{4-x}{(x-1)(x-4)} + \frac{2(x-4)}{x-1(x-4)} =1[/tex]
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Combine the numerator of both terms and bring the denominator to the other side :
[tex]\frac{4-x+2(x-4)}{(x-1)(x-4)} = 1[/tex]
[tex]4 - x+2x-8 = x^{2} - 5x + 4[/tex]
[tex]x - 4 = x^{2} - 5x + 4[/tex]
[tex]x^{2} -5x-x+4+4=0[/tex]
[tex]x^{2} -6x+8=0[/tex]
[tex](x-4)(x-2)=0[/tex]
[tex]x = 2, 4[/tex]
