Vì \(\frac{a}{b}=\frac{c}{d}\) nên ad=bc và \(\frac{a}{c}=\frac{b}{d}=\frac{ab}{cd}\)(1)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{c+d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(2)

Từ (1) và (2), ta suy ra: \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)

Thay:

\(\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)

=> đpcm

Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\)

=> \(\frac{ab}{cd}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)

= \(\frac{ab}{cd}+1-\frac{a^2-b^2}{c^2-d^2}\)

= \(1\)

Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Nên \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

Suy ra : \(\frac{a.b}{c.d}=\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Vậy \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(\text{đ}pcm\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có :

\(\frac{a\times b}{c\times d}=\frac{bk\times b}{dk\times d}=\frac{b^2\times k}{d^2\times k}=\frac{b^2}{d^2}\left(1\right)\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left(b\times\left(k+1\right)\right)^2}{\left(d\times\left(k+1\right)\right)^2}=\frac{b^2\times\left(k+1\right)^2}{d^{2\times}\left(k+1\right)^2}=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) , ta có :\(\frac{a\times b}{c\times d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Đăt \(t=\frac{a}{b}=\frac{c}{d}\)

=> a = b.t; c = d.t

=> \(\frac{a.b}{c.d}=\frac{b.t.b}{d.t.d}=\frac{b^2}{d^2}\)   (1)

Và \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(b.t+b\right)^2}{\left(d.t+d\right)^2}=\frac{b^2\left(t+1\right)^2}{d^2\left(t+1\right)^2}=\frac{b^2}{d^2}\)   (2)

Từ (1) và (2) suy ra điều phải chứng minh

Câu b làm như bạn Thang Tran

Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\)

Đặt \(\frac{b}{a}=\frac{d}{c}=k\)

\(\Rightarrow b=ak\)

\(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+a^2.k^2}{a^2.k^2+c^2}=\frac{a^2}{c^2}=\left(\frac{a}{c}\right)^2=\frac{a}{c}\)