\(e,\Leftrightarrow x^2+6x+9=x^2+4x\\ \Leftrightarrow6x-4x-9\\ \Leftrightarrow2x=-9\Rightarrow x=-\dfrac{9}{2}\\ f,\Leftrightarrow3x-xx-x.1+3x+3.1=6x-x^2\\ \Leftrightarrow5x-6x=-3\\ \Leftrightarrow-x=-3\Leftrightarrow x=3\\ g,\\ \Leftrightarrow\dfrac{x}{4}+\dfrac{x}{3}=2\\ \Leftrightarrow\dfrac{7}{12}x=2\Rightarrow x=\dfrac{24}{7}\\ h,\\ \Leftrightarrow2\left(2x-1\right)-5\left(x+1\right)=0\\ \Leftrightarrow-x-7=0\\ \Rightarrow x=-7\)
e.\(\Leftrightarrow x^2+6x+9=x^2+4x\)
\(\Leftrightarrow2x=-9\Leftrightarrow x=-\dfrac{9}{2}\)
f.\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=x^2-3x\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=x\left(x-3\right)\)
Với \(x=3\) là một nghiệm của pt
Với \(x\ne3\) pt trở thành
\(\Leftrightarrow x+1=x\Leftrightarrow1=0\left(voli\right)\)
Vậy x=3
g.\(\Leftrightarrow\dfrac{3x}{12}=\dfrac{24-4x}{12}\)
\(\Leftrightarrow3x=24-4x\)
\(\Leftrightarrow7x=24\Leftrightarrow x=3\)
h.\(\Leftrightarrow\dfrac{2\left(2x-1\right)-5\left(x+1\right)}{10}=0\)
\(\Leftrightarrow2\left(2x-1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow4x-2-5x-5=0\)
\(\Leftrightarrow x=-7\)
\(e,\left(x+3\right)^2=x^2+4x\\ \Leftrightarrow x^2+6x+9-x^2-4x=0\\ \Leftrightarrow2x=-9\\ \Leftrightarrow x=-\dfrac{9}{2}\\ \Rightarrow S=\left\{-\dfrac{9}{2}\right\}\\ f,3x-\left(x-3\right).\left(x+1\right)=6x-x^2\\ \Leftrightarrow3x-x^2+2x+3-6x+x^2=0\\ \Leftrightarrow-x=-3\\ \Leftrightarrow x=3\\ \Rightarrow S=\left\{3\right\}\\ g,\dfrac{x}{4}=2-\dfrac{x}{3}\\ \Leftrightarrow\dfrac{3x}{12}=\dfrac{24}{12}-\dfrac{4x}{12}\\ \Leftrightarrow3x=24-4x\\ \Leftrightarrow3x+4x=24\\ \Leftrightarrow7x=24\\ \Leftrightarrow x=\dfrac{24}{7}\\ \Rightarrow S=\left\{\dfrac{24}{7}\right\}\\ h,\dfrac{2x-1}{5}-\dfrac{x+1}{2}=0\\ \Leftrightarrow\dfrac{2.\left(2x-1\right)-5.\left(x+1\right)}{10}=0\\ \Leftrightarrow4x-2-5x-5=0\\ \Leftrightarrow-x=7\\ \Leftrightarrow x=-7\\ \Rightarrow S=\left\{-7\right\}\)
mọi người giúp mik với ạ
