Rozwiąż równania:a) b).... Question from @Ghoost86

Ghoost86 @Ghoost86

October 2018 1 19 Report

Rozwiąż równania:

a) $\frac{1}{4x+8}=\frac{20x+1}{4x^2-16}-\frac{7-5x}{x^2-4x+4}$

b) $\frac{2}{x^2-4}-\frac{1}{x^2-2x}+\frac{x-4}{x^2+2x}=0$

SyntiaxD

a) x ≠ -2, x ≠ 2

$\frac{1}{4x+8}=\frac{20x+1}{4x^2-16}-\frac{7-5x}{x^2-4x+4} \\ \\\frac{1}{4(x+2)}=\frac{20x+1}{4(x^2-4)}-\frac{7-5x}{(x-2)^2} \\ \\ \frac{1}{4(x+2)}=\frac{20x+1}{4(x+2)(x-2)}-\frac{7-5x}{(x-2)(x-2)} \\ \\ \frac{1}{4(x+2)}=\frac{(20x+1)(x-2)}{4(x+2)(x-2)(x-2)} -\frac{4(x+2)(7-5x)}{4(x+2)(x-2)(x-2)} \\ \\ \frac{1}{4(x+2)}=\frac{(20x+1)(x-2)-4(x+2)(7-5x)}{4(x+2)(x-2)(x-2)} \\ \\ \frac{1}{4(x+2)}=\frac{20x^2-40x+x-2-4(7x-5x^2+14-10x)}{4(x+2)(x-2)(x-2)}$

$\frac{1}{4(x+2)}=\frac{20x^2-39x-2-28x+20x^2-56+40x}{4(x+2)(x-2)(x-2)} \\ \\ \frac{1}{4(x+2)}=\frac{40x^2-27x-58}{4(x+2)(x-2)(x-2)}\\ \\ 4(x+2)(x-2)(x-2)=4(x+2)(40x^2-27x-58) \\ \\ x^2-4x+4=40x^2-27x-58 \\ \\ 39x^2-23x-62=0$

Δ = (-23)² - 4 · 39 · (-62) = 529 + 9672 = 10201, √Δ = 101

x₁ = $\frac{23-101}{2\cdot39}=\frac{-78}{78}=-1$

x₂ = $\frac{23+101}{2\cdot39}=\frac{124}{78}=\frac{62}{39}=1\frac{23}{39}$

Odp. x = -1, x = $1\frac{23}{39}$

b) x ≠ 0 , x ≠ 2, x ≠ -2

$\frac{2}{x^2-4}-\frac{1}{x^2-2x}+\frac{x-4}{x^2+2x}=0 \\ \\ \frac{2}{(x-2)(x+2)}-\frac{1}{x(x-2)}+\frac{x-4}{x(x+2)}=0 \\ \\ \frac{2x}{x(x-2)(x+2)}+\frac{(x-4)(x-2)}{x(x-2)(x+2)}=\frac{1}{x(x-2)} \\ \\ \frac{x^2-4x+8}{x(x-2)(x+2)}=\frac{1}{x(x-2)} \\ \\ x(x^2-4x+8)(x-2)=x(x-2)(x+2) \\ \\ x^2-5x+6=0$