a: Xét ΔACF và ΔAED có 

AC=AE

\(\widehat{A}\) chung

AF=AD

Do đó: ΔACF=ΔAED

Bài 2:

a) \(\dfrac{2}{15}-\dfrac{7}{10}=\dfrac{4}{30}-\dfrac{21}{30}=-\dfrac{17}{30}\)

b) \(\dfrac{-3}{14}+\dfrac{2}{21}=\dfrac{-9}{42}+\dfrac{4}{42}=\dfrac{-5}{42}\)

c) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-96}{144}+\dfrac{-108}{144}=\dfrac{-204}{144}=-\dfrac{17}{12}\)

Bài 3: 

a) \(\dfrac{3}{8}+\dfrac{-5}{6}=\dfrac{3}{8}-\dfrac{5}{6}=\dfrac{18}{48}-\dfrac{40}{48}=-\dfrac{22}{48}=-\dfrac{11}{24}\)

b) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-24}{54}-\dfrac{30}{54}=\dfrac{-54}{54}=-1\)

c) \(\dfrac{2}{21}-\dfrac{-1}{28}=\dfrac{8}{84}-\dfrac{-3}{84}=\dfrac{11}{84}\)

\(5,=\dfrac{2^3\cdot3^3\cdot2^5\cdot3^7}{2^8\cdot3^9}=3\\ 6,2^{50}+2^{49}-2^{48}=2^{46}\left(2^4+2^3-2^2\right)=2^{46}\cdot20⋮5\left(20⋮5\right)\)

a. \(PTHH:3H_2+Fe_2O_3\underrightarrow{t^o}2Fe+3H_2O\)

b. \(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{4}{160}=0,025\left(mol\right)\)

\(PTHH:3H_2+Fe_2O_3\underrightarrow{t^o}2Fe+3H_2O\)

Mol :         3     :     1      :     2     :      3

Mol :     0,075 ← 0,025 → 0,05  →  0,075

\(\Rightarrow n_{H_2}=0,075\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,075.22,4=1,68\left(l\right)\)

c. Từ câu b. \(\Rightarrow n_{Fe}=0,05\left(mol\right)\)

\(\Rightarrow m_{Fe}=n_{Fe}.M_{Fe}=0,05.56=2,8\left(g\right)\)