Question

giải pt theo cách tính \(\Delta\)

a, \(4x^2+3=\dfrac{3}{x^2}+2\)

b, \(x^4-3=\left(x+1\right)\left(x-1\right)\)

c,\(\dfrac{1}{x-2}=3+\dfrac{2}{x-4}\)

Answer 1

b) x⁴ - 3 = (x + 1)(x - 1)

\(\Rightarrow\) x⁴ - 3 = x² - 1

\(\Rightarrow\) x⁴ = x² + 2

\(\Rightarrow\) 2 = x⁴ - x²

\(\Rightarrow\) 2,25 = (x² - 0,5)²

\(\Rightarrow\) \(\left[{}\begin{matrix}x^2-0,5=1,5\\x^2-0,5=-1,5\end{matrix}\right.\) mà x² - 0,5 \(\ge\) -0,5 nên x² - 0,5 = 1,5

\(\Rightarrow x^2=2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)