\(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\)
\(\Leftrightarrow\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\)
\(\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\)
\(\Leftrightarrow20+40a=105-45a\)
\(\Leftrightarrow40a+20-105+45a=0\)
\(\Leftrightarrow85a-85=0\)
\(\Leftrightarrow a=1\)
Thay a = 1 vào, ta được:
\(\dfrac{7-3}{20}=\dfrac{3b}{23+7}\)
\(\Leftrightarrow\dfrac{1}{5}=\dfrac{3b}{30}\)
\(\Leftrightarrow b=2\)
Vậy a = 1 ; b = 2.
Đúng 0
Bình luận (0)
