Câu 13: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(bk\right)^{2017}+\left(dk\right)^{2017}}{b^{2017}+d^{2017}}=\frac{k^{2017}\left(b^{2017}+d^{2017}\right)}{b^{2017}+d^{2017}}=k^{2017}\)
\(\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\frac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)
Do đó: \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
Câu 12:
a: Ax//DE
=>\(\hat{xAB}+\hat{ABE}=180^0\) (hai góc trong cùng phía)
=>\(\hat{ABE}=180^0-145^0=35^0\)
b: Ta có: \(\hat{DBC}+\hat{BCy}=125^0+55^0=180^0\)
mà hai góc này là hai góc ở vị trí trong cùng phía
nên BD//Cy
=>Cy//DE
mà DE//Ax
nên Cy//Ax
c: BE//Cy
=>\(\hat{EBC}=\hat{BCy}\) (hai góc so le trong)
=>\(\hat{EBC}=55^0\)
Ta có: tia BE nằm giữa hai tia BA và BC
=>\(\hat{ABC}=\hat{ABE}+\hat{CBE}=55^0+35^0=90^0\)
=>BA⊥BC
