Đáp án:
Áp dụng tính chất dãy tỉ số bằng nhau ta có
$\begin{array}{l}
B1)\\
a:b:c = 3:2:5\\
\Leftrightarrow \dfrac{a}{3} = \dfrac{b}{2} = \dfrac{c}{5} = \dfrac{{a - c}}{{3 - 5}} = \dfrac{{18}}{{ - 2}} = - 9\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 3.\left( { - 9} \right) = - 27\\
b = 2.\left( { - 9} \right) = - 18\\
c = 5.\left( { - 9} \right) = - 45
\end{array} \right.\\
Vậy\,a = - 27;b = - 18;c = - 45\\
B2)\\
2.x = 5.y = 7z\\
\Leftrightarrow \dfrac{{2x}}{{70}} = \dfrac{{5y}}{{70}} = \dfrac{{7z}}{{70}}\\
\Leftrightarrow \dfrac{x}{{35}} = \dfrac{y}{{14}} = \dfrac{z}{{10}} = \dfrac{{x + y + z}}{{35 + 14 + 10}} = \dfrac{{590}}{{50}} = 10\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 350\\
y = 140\\
z = 100
\end{array} \right.\\
Vậy\,x = 350;y = 140;z = 100\\
B3)\\
\widehat A + \widehat B + \widehat C = {180^0}\\
\widehat A:4 = \widehat B:5 = \widehat C:9\\
\Leftrightarrow \dfrac{{\widehat A}}{4} = \dfrac{{\widehat B}}{5} = \dfrac{{\widehat C}}{9} = \dfrac{{\widehat A + \widehat B + \widehat C}}{{4 + 5 + 9}} = \dfrac{{{{180}^0}}}{{18}} = {10^0}\\
\Leftrightarrow \left\{ \begin{array}{l}
\widehat A = {40^0}\\
\widehat B = {50^0}\\
\widehat C = {90^0}
\end{array} \right.\\
Vậy\,\widehat A = {40^0};\widehat B = {50^0};\widehat C = {90^0}
\end{array}$
