Question

[jenny]

giải pt

a).4x²+\(\dfrac{1}{x^2}\)+\(\left|2x-\dfrac{1}{x}\right|-6=0\)

b)\(4x^4+\left|2x^3-x\right|6x^2+1=0\)

^{c)\(\dfrac{x^2-1}{\left|x\right|\left(x-2\right)}=2\)}

Answer 1

[Rimuru tempest]

a) Đặt \(t=\left|2x-\dfrac{1}{x}\right|\Leftrightarrow t^2=\left(2x-\dfrac{1}{x}\right)^2=4x^2-4+\dfrac{1}{x^2}\Leftrightarrow t^2+4=4x^2+\dfrac{1}{x^2}\) ĐK \(t\ge0\)

từ có ta có pt theo biến t : \(t^2+4+t-6=0\)

\(\Leftrightarrow t^2+t-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|2x-\dfrac{1}{x}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{x}=1\\2x-\dfrac{1}{x}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

c: TH1: x>0

Pt sẽ là \(\dfrac{x^2-1}{x\left(x-2\right)}=2\)

=>2x^2-4x=x^2-1

=>x^2-4x+1=0

hay \(x=2\pm\sqrt{3}\)

TH2: x<0

Pt sẽ là \(\dfrac{x^2-1}{-x\left(x-2\right)}=2\)

=>-2x(x-2)=x^2-1

=>-2x^2+4x=x^2-1

=>-3x^2+4x+1=0

hay \(x=\dfrac{2-\sqrt{7}}{3}\)

b:

TH1: 2x^3-x>=0

 \(4x^4+6x^2\left(2x^3-x\right)+1=0\)

=>4x^4+12x^5-6x^3+1=0

\(\Leftrightarrow x\simeq-0.95\left(loại\right)\)

TH2: 2x^3-x<0

Pt sẽ là \(4x^4+6x^2\left(x-2x^3\right)+1=0\)

=>4x^4+6x^3-12x^5+1=0

=>x=0,95(loại)