Câu 7 :
\(a,2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
\(b,\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{3}{2}\right\}\)
\(c,\dfrac{x+1}{x-3}-\dfrac{3}{x+3}=\dfrac{x^2+14}{x^2-9}\left(đkxđ:x\ne\pm3\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-3\left(x-3\right)=x^2-14\)
\(\Leftrightarrow x^2+3x+x+3-3x+9=x^2-14\)
\(\Leftrightarrow x^2-x^2+3x+x-3x+3+9+14=0\)
\(\Leftrightarrow x+26=0\)
\(\Leftrightarrow x=-26\left(nhận\right)\)
Vậy \(S=\left\{-26\right\}\)
Câu 9 :
a, Xét ΔAMN và ΔABC có :
\(\dfrac{AM}{AB}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\dfrac{AM}{AC}=\dfrac{3,5}{7}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{AM}{AB}=\dfrac{AM}{AC}\left(=\dfrac{1}{2}\right)\)
Mà \(\widehat{A}:chung\)
\(\Rightarrow\Delta ANM\sim\Delta ABC\left(c-g-c\right)\)
b, Ta có :\(\Delta ANM\sim\Delta ABC\left(cmt\right)\)
\(\Rightarrow\dfrac{AN}{AB}=\dfrac{MN}{BC}\)
hay \(\dfrac{3}{6}=\dfrac{MN}{8}\)
\(\Rightarrow MN=\dfrac{3.8}{6}=4\left(cm\right)\)
