2. Ta thấy:[tex]x^3-x=(x-1)x(x+1)\vdots 6[/tex]
[tex]\Rightarrow \left\{\begin{matrix} a^3-a\vdots 6\\ b^3-b\vdots 6\\ c^3-c\vdots 6 \end{matrix}\right.\Rightarrow a^3+b^3+c^3-(a+b+c)\vdots 6[/tex]
Mà [tex]a+b+c=1890^{2019}\vdots 6\Rightarrow a^3+b^3+c^3\vdots 6\Rightarrow \frac{a^3+b^3+c^3}{6}\in \mathbb{Z}[/tex]
1. a)Ta có:[tex]\left\{\begin{matrix} \widehat{HAF}+\widehat{AFH}=90^o\\ \widehat{AFH}+\widehat{ABF}=90^o \end{matrix}\right.\Rightarrow \widehat{HAF}=\widehat{ABF}[/tex]
Xét [tex]\Delta ABF và \Delta DAM[/tex]
[tex]\left.\begin{matrix} \widehat{BAF}=\widehat{ADM}=90^o\\ AB=AD\\ \widehat{ABF}=\widehat{DAM} \end{matrix}\right\}\Rightarrow \Delta ABF=\Delta DAM\Rightarrow DM=AF=AE\Rightarrow AEDH là hình chữ nhật[/tex]
b) Xét [tex]\Delta AFH và \Delta BAH[/tex]
[tex]\left.\begin{matrix} \widehat{FAH}=\widehat{ABH}\\ \widehat{AHF}=\widehat{BHA}=90^o \end{matrix}\right\}\Rightarrow \Delta AFH \sim \Delta BAH\Rightarrow \frac{AH}{AF}=\frac{BH}{BA}\Rightarrow \frac{AH}{AE}=\frac{BH}{BC}[/tex]
Lại có:[tex]\left.\begin{matrix} \widehat{HAE}+\widehat{ABH}=90^o\\ \widehat{ABH}+\widehat{HBC}=90^o \end{matrix}\right\}\Rightarrow \widehat{HAE}=\widehat{HBC}[/tex]
Xét [tex]\Delta AHE và \Delta BHC[/tex]
[tex]\left.\begin{matrix} \widehat{HAE}=\widehat{HBC}\\ \frac{AH}{AE}=\frac{BH}{BC} \end{matrix}\right\}\Rightarrow \Delta AHE\sim \Delta BHC\Rightarrow \frac{S_{AHE}}{S_{BHC}}=\frac{1}{4}=(\frac{AE}{BC})^2\Rightarrow \frac{AE}{AH}=\frac{1}{2}[/tex]
Tam giác ABD có AB = AD, AE = AF [tex]\Rightarrow \frac{AE}{AB}=\frac{AF}{AC}\Rightarrow EF//BD[/tex]
Áp dụng định lí Ta-lét cho tam giác ABD có EF // BD ta có:[tex]\frac{EF}{BD}=\frac{AE}{AB}=\frac{1}{2}\Rightarrow BD=2EF\Rightarrow AC=2EF[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
