a) \(x-3=0\)
\(\Leftrightarrow\) \(x=0+3\)
\(\Leftrightarrow\) \(x=3\)
b) \(2x+6=0\)
\(\Leftrightarrow\) \(2x=0-6\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
c) \(2x+3=x+9\)
\(\Leftrightarrow2x+3-x=9\)
\(\Leftrightarrow x+3=9\)
\(\Leftrightarrow x=9-3\)
\(\Leftrightarrow x=6\)
d) \(\left(x-4\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\2x+4+\left(-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Bài nhiều quá nên tôi chỉ làm bài 1, 2 thôi nhé :v
1.
\(a,x-3=0\)
\(\Leftrightarrow x=3\)
Vậy : S = {3}
\(b,2x+6=0\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
Vậy : S = {-3}
\(c,2x+3=x+9\)
\(\Leftrightarrow x=6\)
Vậy : S = {6]
\(d,\left(x+4\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy : S = {-4; -2}
2.
\(a,\dfrac{3x+1}{4}=\dfrac{x+2}{3}\)
\(\Leftrightarrow3\left(3x+1\right)=4\left(x+2\right)\)
\(\Leftrightarrow9x+3=4x+8\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\)
Vậy : S = {1}
\(b,\dfrac{4}{x}-\dfrac{5}{x+1}=0\) (ĐKXĐ : x ≠ 0; x ≠ -1)
\(\Rightarrow4\left(x+1\right)-5x=0\)
\(\Leftrightarrow4x+4-5x=0\)
\(\Leftrightarrow-x=-4\)
\(\Leftrightarrow x=4\) (N)
Vậy : S = {4}
