a) \(\Rightarrow x^2+x-6=x^2-4x\)
\(\Rightarrow5x=6\Rightarrow x=\dfrac{6}{5}\)
b) \(\Rightarrow x^2+6x+9=x^2-25\)
\(\Rightarrow6x=-34\Rightarrow x=-\dfrac{17}{3}\)
c) \(\Rightarrow2x^2-5x-3=x^2+4x+4-9x+2\)
\(\Rightarrow x^2=9\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow x^3-8=19\)
\(\Rightarrow x^3=27\Rightarrow x=3\)
f) \(\Rightarrow25-4x^2-2x=x^2-2x+1+28\)
\(\Rightarrow5x^2=-4\left(VLý\right)\)
Vậy \(S=\varnothing\)
a: Ta có: \(\left(x-2\right)\left(x+3\right)=x\left(x-4\right)\)
\(\Leftrightarrow x^2+x-6-x^2+4x=0\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
b: Ta có: \(\left(x+3\right)^2=\left(x-5\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+6x+9-x^2+25=0\)
hay \(x=-\dfrac{17}{3}\)
