Question

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Answer 1

Ответ:

2) $\frac{4}{x+5} -\frac{3}{x-1} =\frac{26}{x^2+4x-5} -1$

$x\neq -5, x\neq 1$

$\frac{4}{x+5} -\frac{3}{x-1} -\frac{26}{x^2+4x-5} +1=0$

$\frac{4}{x+5} -\frac{3}{x-1} -\frac{26}{x^2+5x-x-5} +1=0$

$\frac{4}{x-5} -\frac{3}{x-1} -\frac{26}{x*(x-5)-(x-5)} +1=0$

$\frac{4}{x+5} -\frac{3}{x-1} -\frac{26}{(x+5)*(x-1)} +1=0$

$\frac{4(x-1)-3(x+5)-26+(x+5)*(x-1)}{(x+5)*(x-1)} =0$

$\frac{4x-4-3x-15-26+x^2-x+5x-5}{(x+5)*(x-1)}=0$

$\frac{-50+x^2+5x}{(x+5)*(x-1)} =0$

$-50+x^2+5x=0$

$5x-50+x^2=0$

$10x-5x-50+x^2=0$

$x*(x+10)-5(x+10)=0$

$(x+10)*(x-5)=0$

$x+10=0$

$x-5=0$

x₁ = -10, x₂ = 5

3)  $\frac{x^2+6}{x} -\frac{5x}{x^2+6}=4$

$x\neq 0$

$\frac{x^2+6}{x} -\frac{5x}{x^2+6} -4=0$

$\frac{(x^2+6)^2-5x^2-4x*(x^2+6)}{x*(x^2+6)} =0$

$\frac{x^4+12x^2+36-5x^2-4x^3-24x}{x*(x^2+6)} =0$

$\frac{x^4+7x^2+36-4x^3-24x}{x*(x^2+6)} =0$

$x^4+7x^2+36-4x^3-24x=0$

$x^4-4x^3+7x^2-24x+36=0$

$x^4-2x^3-2x^3+4x^2+3x^2-6x-18x+36=0$

$x^3*(x-2)-2x^2*(x-2)+3x*(x-2)-18(x-2)=0$

$(x-2)*(x^3-2x^2+3x-18)=0$

$(x-2)*(x^3-3x^2+x^2-3x+6x-18)=0$

$(x-2)*(x^2*(x-3)+x*(x-3)+6(x-3))=0$

$(x-2)*(x-3)*(x^2+x+6)=0$

$x-2=0$

$x-3=0$

$x^2+x+6=0$

x₁ = 2, x₂ = 3