Question

Giải phương trình\(\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{18}\)

Answer 1

ĐKXĐ : x khác -3;-4;-5;-6

pt <=> 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/18

<=> 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/18

<=> 1/x+3 - 1/x+6 = 1/18

<=> x+6-x-3/(x+3).(x+6) = 1/18

<=> 3/x^2+9x+18 = 1/18

<=> x^2+9x+18 = 3 : 1/18 = 48

<=> x^2+9x+18-48 = 0

<=> x^2+9x-30 = 0

<=>(x+9/2)^2 = 201/4

<=> x = \(\frac{+-\sqrt{201}-9}{2}\)(tm)

k mk nha

Answer 2

                   \(\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+6}=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{3}{\left(x+3\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\)(x + 3)(x + 6) = 54 = 6.9 = (-6).(-9)

Đến đây giải tiếp nha