\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)mà\(\frac{a^2}{c^2}=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

?????

og hỏi cái j zạy:D??

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

a) Khi đó, ta có:

 +) \(\frac{bk}{b}=k\)

+) \(\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)

=> \(\frac{a}{b}=\frac{a+c}{b+d}\)

b) Ta có:

 +) \(\frac{bk-b}{b}=\frac{b\left(k-1\right)}{b}=k-1\)

 +) \(\frac{dk-d}{d}=\frac{d\left(k-1\right)}{d}=k-1\)

=> \(\frac{a-b}{b}=\frac{c-d}{d}\)

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

Do đó \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\)(1)

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

Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\left(đpcm\right)\)

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

=> \(a=bk\) 

       \(c=dk\) 

Ta có: 

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

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

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

=> đpcm

Cảm ơn bạn nha. Mình tick đúng cho bạn rồi đó.

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)

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

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

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

Chúc bạn học tốt!

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) (k\(\in\)N*)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Thay vào ta có:

\(\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1}\)

\(\frac{c}{c+d}=\frac{dk}{dk+d}=\frac{dk}{d\left(k+1\right)}=\frac{k}{k+1}\)

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

Vậy \(\frac{a}{a+b}=\frac{c}{c+d}\)(điều phải chứng minh)

Hok tốt nha!!!

Theo đề, ta có: 6a=2b=-4c=5d

\(\Leftrightarrow\dfrac{a}{10}=\dfrac{b}{30}=\dfrac{c}{-15}=\dfrac{d}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{10}=\dfrac{b}{30}=\dfrac{c}{-15}=\dfrac{d}{12}=\dfrac{3a-2b+4c-d}{3\cdot10-2\cdot30+4\cdot\left(-15\right)-12}=\dfrac{2}{-102}=-\dfrac{1}{51}\)

Do đó: a=-10/51; b=-10/17; c=5/17; d=4/17

\(a+b-2c-3d=\dfrac{-10}{51}-\dfrac{10}{17}-2\cdot\dfrac{5}{17}-3\cdot\dfrac{4}{17}=-\dfrac{106}{51}\)