b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
\(=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}\)
\(=k^2\)(1)
Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}\)
\(=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}\)
\(=k^2\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k\)
\(\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k\)
Do đó: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}\)
Giúp mình với ạ mình đang cần gấp:((