Xét ΔABC có

\(\widehat{B}+\widehat{C}+\widehat{A}=180^0\)

\(\Leftrightarrow2\cdot\widehat{B}=180^0-\widehat{A}\)

hay \(\widehat{B}=\widehat{C}=\dfrac{180^0-\widehat{A}}{2}\)

Hông bé ơi 😊🤣🤣🤣🤣😅😂😂

a: =>x+2/3=4/60=1/15

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

b: =>x/3=14/21-3/21=11/21

=>x=11/7

\(x+\dfrac{2}{3}=-\dfrac{1}{12}.\dfrac{-4}{5}.\\ \Leftrightarrow x+\dfrac{2}{3}=\dfrac{1}{15}.\\ \Leftrightarrow x=\dfrac{-3}{5}.\\ \dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\\Leftrightarrow \dfrac{7x}{21}=\dfrac{11}{21}.\\\Leftrightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}. \)

\(a,x+\dfrac{2}{3}=-\dfrac{1}{12}.\dfrac{-4}{5}=\dfrac{1}{15}\)

\(x=\dfrac{1}{15}-\dfrac{2}{3}=-\dfrac{3}{5}\)

\(b,\dfrac{x}{3}=\dfrac{2}{3}+\left(-\dfrac{1}{7}\right)=\dfrac{11}{21}\)

\(\dfrac{7x}{21}=\dfrac{11}{21}\)

\(x=\dfrac{11}{7}\)

14.8 và câu14.9 thôi nhé

a)

b)\(U_{12}=3,2V;U_{23}=5,6V\)

\(Đ_1ntĐ_2\Rightarrow U=U_1+U_2=3,2+5,6=8,8V\)

c)\(I_1=I_2=I_m=0,35A\)