\(9x^4-10x^2+1=0\\ \Rightarrow\left(9x^4-9x^2\right)-\left(x^2-1\right)=0\\ \Rightarrow9x^2\left(x^2-1\right)-\left(x^2-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(9x^2-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(3x-1\right)\left(3x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Đặt x^2 = t ( t>= 0 )
9t^2 - 10t + 1 = 0
ta có : a + b + c = 9 - 10 + 1 = 0
=> t = 1 ; t = 1/9
theo cách đặt x = 1 ; x = 1/3
\(\Leftrightarrow\left(9x^2-1\right)\left(x^2-1\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};1;-1\right\}\)
gọi x2 là t:
9t2-10t+1=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
ta theo cách đặt của pt ta có: x=1
x=\(\dfrac{1}{3}\)
