a) Ta có\(\frac{3a-b}{3a+b}=\frac{3c-d}{3c+d}\)

=> (3a - b)(3c + d) = (3a + b)(3c - d)

=> 9ac + 3ad - 3bc - bd = 9ac - 3ad + 3bc - bd

=> 3ad - 3bc = -3ad + 3bc

=> 3ad + 3ad = 3bc + 3bc

=> 6ad = 6bc

=> ad = bc

=> \(\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)

b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

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

\(\frac{bd}{ac}=\frac{bd}{bkdk}=\frac{1}{k^2}\left(2\right)\)

Từ (1)(2) => \(\frac{b^2+d^2}{a^2+c^2}=\frac{bd}{ac}\)(đpcm)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)

Ta có:\(\frac{a^2+ac}{c^2-ac}=\frac{b^2k^2+bk.dk}{d^2k^2-bk.dk}=\frac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(1)

\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(2)

Từ 1 và 2 =>\(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
 

What là cái gì?

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)

Ta có: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2.k^2+bk.dk}{d^2.k^2-bk.dk}=\frac{bk^2.\left(b+d\right)}{dk^2.\left(d-b\right)}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (1)

\(\frac{b^2+bd}{d^2-bd}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (2)

Từ (1) và (2) => \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)

Chúc bạn học tốt!

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk\cdot dk}{\left(dk\right)^2-bk\cdot dk}=\frac{b^2k^2+bd\cdot k^2}{d^2k^2-bd\cdot k^2}=\frac{b\cdot k^2\left(b+d\right)}{d\cdot k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

a) \(VT=\frac{a}{a+c}=\frac{kb}{kb+kd}=\frac{kb}{k\left(b+d\right)}=\frac{b}{b+d}=VP\)

=> đpcm

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

\(VP=\frac{ac}{bd}=\frac{kb\cdot kd}{bd}=\frac{k^2bd}{bd}=k^2\)(2)

Từ (1) và (2) => VT = VP => đpcm

a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

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

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

Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

b: \(\frac{7a-4b}{3a+5b}=\frac{7\cdot bk-4b}{3\cdot bk+5b}=\frac{b\left(7k-4\right)}{b\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

\(\frac{7c-4d}{3c+5d}=\frac{7\cdot dk-4d}{3\cdot dk+5d}=\frac{d\left(7k-4\right)}{d\left(3k+5\right)}=\frac{7k-4}{3k+5}\)

Do đó: \(\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}\)

c: \(\frac{ac}{bd}=\frac{bk\cdot dk}{bd}=k^2\)

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

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

Do đó; \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

d: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)

\(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)

Do đó: \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(a+b\right)^3}{\left(c+d\right)^3}\)

Do đó:

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{c}{d}.\frac{c}{d}\Leftrightarrow\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}\)

Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

Ta có đpcm

Tham khảo:Chứng minh a/b=c/d hoặc a/b=d/c biết (a^2+b^2)/(c^2+d^2)=ab/cd - An Nhiên

\(\text{Cho }\dfrac{a}{b}=\dfrac{d}{c}\text{ và }b,d\notin0\text{.CMR:}\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

\(\text{Ta có:}\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\text{Lại có:}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{\left(bd\right).k^2}{bd}=k^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{\left(b^2+d^2\right).k^2}{b^2+d^2}=k^2\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)