Question

giải phương trình :

\(\dfrac{1}{5x^2}+\dfrac{1}{x^2-9x+36}=\dfrac{1}{x^2-4x^2+16}\)

Answer 1

chỉnh đề: \(\dfrac{1}{5x^2}+\dfrac{1}{x^2-9x+36}=\dfrac{1}{x^2-4x+16}\)

\(\dfrac{1}{5x^2}+\dfrac{1}{x^2}.\dfrac{1}{1-\dfrac{9}{x}+\dfrac{36}{x^2}}=\dfrac{1}{x^2}.\dfrac{1}{1-\dfrac{4}{x}+\dfrac{16}{x^2}}\)nhân hai vế cho x^2 khác 0

đặt [-1/x +4/x^2 ] =t

<=>\(\dfrac{1}{5}+\dfrac{1}{1+9t}=\dfrac{1}{1+4t}\)

<=>( 9t+1)(4t+1) +5(4t+1) =5(9t +1)

<=>( 4t+1)[(9t+6) =5(9t +1)

<=> 36t^2 -12t +1 =(6t -1)^2 =0

= > t =1/6

[-1/x +4/x^2 ] =1/6

x^2 +6x-24 =0

9+24 =33

\(\left[{}\begin{matrix}x_1=-3-\sqrt{33}\\x_2=-3+\sqrt{33}\end{matrix}\right.\)