\(\dfrac{2x-2}{x+2}+\dfrac{3}{x+2}=\dfrac{x}{x+2}\left(x\ne-2\right)\\ < =>\dfrac{2x+1}{x+2}=\dfrac{x}{x+2}\\ =>2x+1=x\\ < =>2x-x=-1\\ < =>x=-1\left(TMDK\right)=>S=\left\{-1\right\}\)

Ta có \(\dfrac{EA}{EB}=3\Rightarrow\dfrac{EA}{AB}=\dfrac{3}{4}\)

\(\dfrac{FC}{FA}=3\Rightarrow\dfrac{FA}{FC}=\dfrac{1}{3}\Rightarrow\dfrac{FA}{AC}=\dfrac{1}{4}\)

Hai tg AEC và tg ABC có chung đường cao từ C->AB nên

\(\dfrac{S_{AEC}}{S_{ABC}}=\dfrac{EA}{AB}=\dfrac{3}{4}\Rightarrow S_{AEC}=\dfrac{3}{4}xS_{ABC}\)

Hai tg AEF và tg AEC có chung đường cao từ E->AC nên

\(\dfrac{S_{AEF}}{S_{AEC}}=\dfrac{FA}{AC}=\dfrac{1}{4}\Rightarrow S_{AEF}=\dfrac{1}{4}xS_{AEC}=\dfrac{1}{4}x\dfrac{3}{4}xS_{ABC}=\dfrac{3}{16}xS_{ABC}\)

\(\Rightarrow\dfrac{S_{AEF}}{S_{ABC}}=\dfrac{3}{16}\)

s

x -2x = 6+3 

x=9 

\(x-6=3+2x\)

\(\Rightarrow x-2x=3+6\)

\(\Rightarrow x\left(1-2\right)=9\)

\(\Rightarrow x.\left(-1\right)=9\)

\(\Rightarrow x=9:\left(-1\right)\)

\(\Rightarrow x=-9\)

Ta thấy : 

\(x^2-6x+17=\left(x^2-6x+9\right)+8\\ =\left(x-3\right)^2+8\ge8>0\forall x\)

Vậy phương trình vô nghiệm

pt vô nghiệm

a/ Xét tg vuông HAB và tg ABC có

 \(\widehat{ABC}\)  chung

=> tg HAB đồng dạng với tg ABC (g.g.g)

b/ Xét tg vuông HAC và tg vuông ABC có

\(\widehat{ACB}\) chung 

=> tg HAC đồng dạng với tg ABC (g.g.g)

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{HC}{AC}\Rightarrow AC^2=HC.BC\)

a/

\(x^3\left(x-4\right)+x\left(x-4\right)=\left(x-4\right)\left(x^3+x\right)=\)

\(=x\left(x^2+1\right)\left(x-4\right)\)

b/

\(\left(x-1\right)\left(x^2-9\right)-2x\left(x-3\right)=\)

\(=\left(x-1\right)\left(x-3\right)\left(x+3\right)-2x\left(x-3\right)=\)

\(=\left(x-3\right)\left[\left(x-1\right)\left(x+3\right)-2x\right]=\)

\(=\left(x-3\right)\left(x^2-3\right)\)

\(\left(a\right):=x\left(y+z\right)+y\left(y+z\right)=\left(y+z\right)\left(x+y\right)\\ \left(b\right):=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\\ \left(c\right):=4^2-\left(x+y\right)^2=\left(4+x+y\right)\left(4-x-y\right)\\ \left(d\right):=x\left(x+y+z\right)-\left(x+y+z\right)=\left(x+y+z\right)\left(x-1\right)\)