Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)

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

+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)

Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)

+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{2c}{2d}=\dfrac{c}{d}\) (1)

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{3a+3c}{3b+3d}=\dfrac{a+3c-\left(3a+3c\right)}{b+3d-\left(3b+3d\right)}=\dfrac{-2a}{-2b}=\dfrac{a}{b}\) (2)

(1);(2) \(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

Đặt \(P=\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b}\)

\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{ac+cd}+\dfrac{d^2}{ad+bd}\)

\(P\ge\dfrac{\left(a+b+c+d\right)^2}{ab+2ac+bc+2bd+cd+ad}=\dfrac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+ab+bc+cd+ad}\)

\(P\ge\dfrac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=d\)

ca này để thầy lâm ròi:<

\(\left(a^2+b+c+d\right)\left(1+b+c+d\right)\ge\left(a+b+c+d\right)^2=16\)

\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}=\dfrac{5-a}{16}\)

Tương tự: \(\dfrac{1}{b^2+c+d+a}\le\dfrac{5-b}{16}\) ...

Cộng vế:

\(P\le\dfrac{20-\left(a+b+c+d\right)}{16}=1\)

Dấu "=" xảy ra khi \(a=b=c=d=1\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)

\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)

=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)

b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)

=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)

Nhanh nha 

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

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

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

\(\dfrac{a}{b}=\dfrac{c}{d}=>\)\(ad=bc\)

\(\dfrac{a}{a-b}=\dfrac{ad}{d\left(a-b\right)}=\dfrac{bc}{ad-bd}=\dfrac{bc}{bc-bd}=\dfrac{bc}{b\left(c-d\right)}\dfrac{c}{c-d}\)

a) \(\dfrac{a}{b}=\dfrac{c}{d}\left(a;b;c;d\ne0\right)\)

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

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

\(\Rightarrow dpcm\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

\(\Rightarrow dpcm\)

Thanks

Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k$

$\Rightarrow a=bk; c=dk$. Khi đó:

1.

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

$\frac{c+d}{d}=\frac{dk+d}{d}=\frac{d(k+1)}{d}=k+1(2)$
Từ $(1); (2)\Rightarrow \frac{a+b}{b}=\frac{c+d}{d}$

2.

$\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}(3)$

$\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}(4)$

Từ $(3); (4)\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}$ (đpcm)

Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?