Question

a,

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

b,\(3\cdot\left(\dfrac{x+3}{x-2}\right)^2+68\cdot\left(\dfrac{x-3}{x+2}\right)^2-46\cdot\dfrac{x^2-9}{x^2-4}=6\)

Answer 1

a: \(\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}\)

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

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

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

=>x(x-x^2+x-1)=0

=>x(-x^2+2x-1)=0

=>x=0(loại) hoặc x=1(nhận)

b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2

=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)

=>\(x\simeq28,4\)