Đáp án:
Bài 1:
$\begin{array}{l}
a)\text{Xét}:\Delta ABE;\Delta ACE:\\
+ AB = AC\\
+ AE\,chung\\
+ BE = CE\\
\Rightarrow \Delta ABE = \Delta ACE\left( {c - c - c} \right)\\
\text{Vậy}\,\Delta ABE = \Delta ACE\\
b)Do:\Delta ABE = \Delta ACE\\
\Rightarrow \widehat {AEB} = \widehat {AEC}\\
Do:\widehat {AEB} + \widehat {AEC} = {180^0}\\
\Rightarrow \widehat {AEB} = \widehat {AEC} = {90^0}\\
\Rightarrow AE \bot BC\\
c)Do:AB = AC\\
\Rightarrow \widehat {ABC} = \widehat {ACB}\\
\Rightarrow \widehat {ABF} = \widehat {ACG}\\
\text{Xét}:\Delta ABF;\Delta ACG:\\
+ AB = AC\\
+ \widehat {ABF} = \widehat {ACG}\\
+ BF = CG\\
\Rightarrow \Delta ABF = \Delta ACG\left( {c - g - c} \right)\\
\Rightarrow \widehat {AFB} = \widehat {AGC}\\
\text{Vậy}\,\widehat {AFB} = \widehat {AGC}\\
B2)\\
25.{\left( { - \dfrac{1}{3}} \right)^3} + \dfrac{1}{5} - 2.{\left( { - \dfrac{1}{2}} \right)^2} - \dfrac{1}{2}\\
= 25.\dfrac{{ - 1}}{{27}} + \dfrac{1}{5} - 2.\dfrac{1}{2} - \dfrac{1}{2}\\
= \dfrac{{ - 25}}{{27}} + \dfrac{1}{5} - \dfrac{1}{2} - \dfrac{1}{2}\\
= \dfrac{{ - 25}}{{27}} + \dfrac{1}{5} - 1\\
= \dfrac{{ - 25.5 + 27 - 27.5}}{{27.5}}\\
= \dfrac{{ - 125 + 27 - 136}}{{135}}\\
= \dfrac{{ - 26}}{{15}}
\end{array}$
