Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\left(ĐPCM\right)\)

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

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

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

Ta có: 

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

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

\(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)

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

\(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)(Tính chất dãy tỉ số bằng nhau)

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

=> \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(Đpcm)

a)\(\frac{ab}{cd}=\frac{bk.b}{dk.b}=\frac{b^2}{d^2}\left(1\right)\)

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

từ\(\left(1\right)\)và\(\left(2\right)\)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

        Lời giải

Ta có: \(\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a.b}{b.c}=\frac{a}{c}\) (1)

Mặt khác,áp dụng t/c tỉ dãy số bằng nhau,ta có:\(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\) (2)

Từ (1) và (2) ta có đpcm (điều phải chứng minh)

tth, Cảm ơn bạn nhìu!

đặt \(\frac{a}{b}=\frac{b}{c}=k\)=> a= bk; b= ck

xét:

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

+) \(\frac{a}{c}\)=\(\frac{bk}{c}=\frac{ckk}{c}=k^2\)

=> \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)đpcm

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

\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\\ \frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)

Do đó: \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

a= 60

b= 40

c= 84

Cách giải thế nào hả bạn?