1.
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(7a=9b=21c=\dfrac{a}{\dfrac{1}{7}}=\dfrac{b}{\dfrac{1}{9}}=\dfrac{c}{\dfrac{1}{21}}=\dfrac{a-b+c}{\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{21}}=\dfrac{15}{\dfrac{5}{63}}=15\cdot\dfrac{63}{5}=189\\ \Rightarrow\left\{{}\begin{matrix}7a=189\\9b=189\\21c=189\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=189:7\\b=189:9\\c=189:21\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=27\\b=21\\c=9\end{matrix}\right.\)
2.
\(b^2=ac\Rightarrow\dfrac{b}{c}=\dfrac{a}{b}\)
\(\dfrac{b}{c}=\dfrac{a}{b}=k\Rightarrow b=ck;a=bk\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{b^2k^2+c^2k^2}{b^2+c^2}=\dfrac{k^2\left(b^2+c^2\right)}{b^2+c^2}=k^2\\ \dfrac{a}{c}=\dfrac{bk}{c}=\dfrac{ck\cdot k}{c}=k^2\\ \Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
Câu 2:
Ta có:
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)
\(\RightarrowĐPCM\)
Câu 1:
7a = 9b = 21c
\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{9}=\dfrac{c}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{21}=\dfrac{b}{9}=\dfrac{c}{7}=\dfrac{a-b+c}{21-9+7}=\dfrac{-15}{19}\)
\(\Rightarrow\left\{{}\begin{matrix}a=-15:19.21\\b=-15:19.9\\c=-15:19.7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-16\dfrac{11}{19}\\b=-7\dfrac{2}{19}\\c=-5\dfrac{10}{19}\end{matrix}\right.\)
