1/x^2+5x+6 + 1/x^2+7x+12 + 1/x^2+9x+20 + 1/x^2+11x+30 =1/8 22/08/2021 Bởi Sadie 1/x^2+5x+6 + 1/x^2+7x+12 + 1/x^2+9x+20 + 1/x^2+11x+30 =1/8
Đk: x (ko) = 2 ,x (ko) = 3 ,x (ko) = 4 , x (ko) = 5, x (ko) = 6 Pt ⇔ $\frac{1}{x-5.x-6}$ + $\frac{1}{x-5.x-4}$ +…+$\frac{1}{x-3.x-2}$ = $\frac{1}{8}$ ⇔ $\frac{1}{x-6}$ – $\frac{1}{x-5}$ + $\frac{1}{x-5}$ – $\frac{1}{x-4}$ + $\frac{1}{x-4}$ +…+ $\frac{1}{x-3}$ – $\frac{1}{x-2}$ = $\frac{1}{8}$ ⇔ $\frac{1}{x-6}$ – $\frac{1}{x-2}$ = $\frac{1}{8}$ ⇔ $\frac{x-2}{x-6.x-2}$ – $\frac{x-6}{x-2.x-6}$=$\frac{1}{8}$ ⇔$\frac{4}{x-6.x-2}$ = $\frac{1}{8}$ ⇔ (x-6).(x-2) = 32 ⇔ x² – 8x + 12 = 32 ⇔ x² – 8x – 20 = 0 ⇔ (x – 10).(x+2) = 0 ⇔ \(\left[ \begin{array}{l}x=10\\x=-2\end{array} \right.\) Bình luận
Đk: x (ko) = 2 ,x (ko) = 3 ,x (ko) = 4 , x (ko) = 5, x (ko) = 6
Pt ⇔ $\frac{1}{x-5.x-6}$ + $\frac{1}{x-5.x-4}$ +…+$\frac{1}{x-3.x-2}$ = $\frac{1}{8}$
⇔ $\frac{1}{x-6}$ – $\frac{1}{x-5}$ + $\frac{1}{x-5}$ – $\frac{1}{x-4}$ + $\frac{1}{x-4}$ +…+ $\frac{1}{x-3}$ – $\frac{1}{x-2}$ = $\frac{1}{8}$
⇔ $\frac{1}{x-6}$ – $\frac{1}{x-2}$ = $\frac{1}{8}$ ⇔ $\frac{x-2}{x-6.x-2}$ – $\frac{x-6}{x-2.x-6}$=$\frac{1}{8}$
⇔$\frac{4}{x-6.x-2}$ = $\frac{1}{8}$ ⇔ (x-6).(x-2) = 32
⇔ x² – 8x + 12 = 32 ⇔ x² – 8x – 20 = 0
⇔ (x – 10).(x+2) = 0 ⇔ \(\left[ \begin{array}{l}x=10\\x=-2\end{array} \right.\)