Question

Giúp mình bài này nhé Tìm x a) ( 4x^4 + 3x^3 ) : ( - x^3 ) + ( 15 x^3 + 6x ) : 3x =0 b) ( x^2 - 1/2x ) : 2x - ( 3x - 1)^2 : ( 3x - 1) = 0 Cứu mình với

Answer 1

a: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^3+6x}{3x}=0\)

\(\Leftrightarrow-4x-3+5x^2+2=0\)

\(\Leftrightarrow5x^2-4x-1=0\)

\(\Leftrightarrow5x^2-5x+x-1=0\)

=>(x-1)(5x+1)=0

=>x=1 hoặc x=-1/5

b: \(\dfrac{x^2-\dfrac{1}{2}x}{2x}-\dfrac{\left(3x-1\right)^2}{3x-1}=0\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-3x+1=0\)

\(\Leftrightarrow\dfrac{-5}{2}x+\dfrac{3}{4}=0\)

\(\Leftrightarrow-\dfrac{5}{2}x=-\dfrac{3}{4}\)

hay \(x=\dfrac{3}{4}:\dfrac{5}{2}=\dfrac{3}{4}\cdot\dfrac{2}{5}=\dfrac{6}{20}=\dfrac{3}{10}\)