Câu 9:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=b\cdot k;c=d\cdot k\)
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Câu 10:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{a-2b}{b}=\dfrac{bk-2b}{b}=\dfrac{b\left(k-2\right)}{b}=k-2\)
\(\dfrac{c-2d}{d}=\dfrac{dk-2d}{d}=\dfrac{d\left(k-2\right)}{d}=k-2\)
Do đó: \(\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\)
Câu 11:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\cdot\left(bk\right)^2+3\cdot bk\cdot b}{11\cdot\left(bk\right)^2-8b^2}\)
\(=\dfrac{7b^2k^2+3kb^2}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\cdot\left(dk\right)^2+3\cdot dk\cdot d}{11\cdot\left(dk\right)^2-8\cdot d^2}\)
\(=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)