Ta có

\(\widehat{A}+\widehat{D}=180^o\) (Hai dt // bị cắt bởi 1 đường thẳng tạo thành 2 góc trong cùng phía bù nhau)

\(\widehat{DAE}=\frac{\widehat{A}}{2}\)

\(\widehat{ADE}=\frac{\widehat{D}}{2}\)

\(\Rightarrow\widehat{DAE}+\widehat{ADE}=\frac{\widehat{A}+\widehat{D}}{2}=\frac{180^o}{2}=90^o\)

Xét tg AED có

\(\widehat{DAE}+\widehat{ADE}=90^o\Rightarrow\widehat{AED}=90^o\Rightarrow AE\perp DE\)

a) \(x^2+8x+15=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\)

b) \(x^2+3x+2=x^2+2x+x+2=\left(x+1\right)\left(x+2\right)\)

c) \(-x^2+7x-6=-x^2+x+6x-6=\left(-x+6\right)\left(x-1\right)\)

d) \(5x^3y-10x^2y^2+5xy^3=5xy\left(x^2-2xy+y^2\right)=5xy\left(x-y\right)^2\)

e) \(2x^2+7x-15=2x^2-3x+10x-15=\left(2x-3\right)\left(x+5\right)\)

\(ax^2+a-axy+2ax-ay\)

\(a\left(x^2+2x+1\right)-ay\left(x+1\right)\)

\(a\left(x+1\right)^2-ay\left(x+1\right)\)

\(\left(x+1\right)\left[a\left(x+1\right)-ay\right]\)

\(\left(x+1\right)\left(ax+a-ay\right)\)

\(a\left(x+1\right)\left(x+1-y\right)\)

Ta xét VT:

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)

Ta xét VP:

\(\left(a-b\right)\left(b-c\right)\left(a-c\right)=\left(ab-ac-b^2+bc\right)\left(a-c\right)\)

\(=a^2b-a^2c-ab^2+abc-abc+ac^2+b^2c-bc^2\)

\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)

Ta thấy: VT = VP 

\(\Rightarrowđpcm\)