Question

giải phương trình

8(x+\(\dfrac{1}{x}\))\(^2\) +4(x\(^2\) +\(\dfrac{1}{x^2}\))\(^2\) -4(x\(^2\) + \(\dfrac{1}{x^2}\))(x+\(\dfrac{1}{x}\))\(^2\)=(x+4)\(^2\)

Answer 1

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4.\left(x^2+\dfrac{1}{x^2}\right)\left[-2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left[\left(x+\dfrac{1}{x}\right)^2-\left(x^2+\dfrac{1}{x^2}\right)\right]=\left(x+4\right)^2\)

\(x\ne0\Leftrightarrow\left(x+4\right)^2=16\Rightarrow\left[{}\begin{matrix}x+4=4;x=0\left(l\right)\\x+4=-4x;x=-8\end{matrix}\right.\)

Answer 2

x+1/x =a

x^2 +1/x^2 =a-2

<=> 8a^2 +4(a^2 -2)^2 -4(a^2 -2) a^2 =(x+4)^2

<=> 8a^2 +4 a^4 -16a^2 +16-4 a^4 + 8 a^2 =(x+4)^2

<=> 8a^2 +4 a^4 -16a^2 +16-4 a^4 +8 a^2 =(x+4)^2

<=> 16 =(x+4)^2

x+4 =4 => x =8

x+4 =-4 => x=0

Answer 3

x khac 0

a =x+1/x => |a| >=2

x^2 +1/x^2 =a-2

<=> 8a^2 +4(a^2 -2)^2 -4(a^2 -2) a^2 =(x+4)^2

<=> 8a^2 +4 a^4 -16a^2 +16-4 a^4 + 8 a^2 =(x+4)^2

<=> 8a^2 +4 a^4 -16a^2 +16-4 a^4 +8 a^2 =(x+4)^2

<=> 16 =(x+4)^2

x+4 =4 => x =8