Bài 1:
Đặt 2x+1=a
Theo đề, ta có: \(\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}=3\)
=>3a^2(a+1)^2=a^2+2a+1+a^2
=>3a^2(a^2+2a+1)-2a^2-2a-1=0
=>3a^4+6a^3+a^2-2a-1=0
=>(a^2+a-1)(3a^2+3a+1)=0
=>\(a\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)
=>\(2x+1\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)
=>\(2x\in\left\{\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)
hay \(x\in\left\{\dfrac{-3+\sqrt{5}}{4};\dfrac{-3-\sqrt{5}}{4}\right\}\)
Đúng 0
Bình luận (0)
