Bài 13:
Theo hình vẽ, ta có: AG=GH=HI=IB=AB/4 và AF=FE=ED=DC=AC/4
Xét ΔABC có \(\dfrac{AG}{AB}=\dfrac{AF}{AC}=\dfrac{1}{4}\)
nên GF//BC
Xét ΔABC có GF//BC
nên \(\dfrac{GF}{BC}=\dfrac{AG}{AB}\)
=>\(\dfrac{GF}{120}=\dfrac{1}{4}\)
=>\(GF=120\cdot\dfrac{1}{4}=30\left(m\right)\)
AG+GH=AH
=>\(AH=\dfrac{1}{4}AB+\dfrac{1}{4}AB=\dfrac{1}{2}AB\)
AF+FE=AE
=>\(AE=\dfrac{1}{4}AC+\dfrac{1}{4}AC=\dfrac{1}{2}AC\)
Xét ΔABC có \(\dfrac{AH}{AB}=\dfrac{AE}{AC}\left(=\dfrac{1}{2}\right)\)
nên HE//BC
Xét ΔABC có HE//BC
nên \(\dfrac{HE}{BC}=\dfrac{AH}{AB}\)
=>\(\dfrac{HE}{120}=\dfrac{1}{2}\)
=>\(HE=60\left(m\right)\)
AH+HI=AI
=>\(AI=\dfrac{1}{2}AB+\dfrac{1}{4}AB=\dfrac{3}{4}AB\)
AE+ED=AD
=>\(AD=\dfrac{1}{2}AC+\dfrac{1}{4}AC=\dfrac{3}{4}AC\)
Xét ΔABC có \(\dfrac{AI}{AB}=\dfrac{AD}{AC}\left(=\dfrac{3}{4}\right)\)
nên ID//BC
Xét ΔABC có ID//BC
nên \(\dfrac{ID}{BC}=\dfrac{AI}{AB}\)
=>\(\dfrac{ID}{120}=\dfrac{3}{4}\)
=>\(ID=120\cdot\dfrac{3}{4}=90\left(m\right)\)
Bài 15:
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
=1
