Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\left(1\right)\)
a) Thay (1) vào đề:
\(VT=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)
\(VP=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\)
\(\Rightarrow VT=VP\)
\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)
b) Thay (1) vào đề bài:
\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)
Theo câu a) \(\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
a) Ta có : \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{k^2}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
