Question

giúp mik vs đang cần gấp

Answer 1

\(1)15-3x=0.\\ \Leftrightarrow3x=15.\\ \Leftrightarrow x=5.\)

\(2)4x+12=2x\left(x+3\right).\\ \Leftrightarrow4\left(x+3\right)-2x\left(x+3\right)=0.\\ \Leftrightarrow\left(4-2x\right)\left(x+3\right)=0.\) 

\(\Leftrightarrow\left[{}\begin{matrix}4-2x=0.\\x+3=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-3.\end{matrix}\right.\)

\(3)\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}.\left(x\ne1\right).\\ \Leftrightarrow\dfrac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0.\\ \Rightarrow3x^2-3x=0.\\ \Leftrightarrow3x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right).\\x=1\left(koTM\right).\end{matrix}\right.\)

 

Answer 2

1A

2B =)