Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

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

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

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

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

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

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

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

Đặt :

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

Ta có :

\(\dfrac{2d-3c}{d}=\dfrac{2d-3dk}{d}=\dfrac{d\left(2-3k\right)}{d}=2-3k\left(1\right)\)

\(\dfrac{2b-3a}{b}=\dfrac{2b-3bk}{b}=\dfrac{b\left(2-3k\right)}{b}=2-3k\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Ta có:(dfrac{a}{b}=dfrac{c}{d}Rightarrowdfrac{a}{c}=dfrac{b}{d}=dfrac{3a}{3c}=dfrac{2b}{2d})

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

(dfrac{b}{d}=dfrac{3a}{3c}=dfrac{2b}{2d}=dfrac{2b-3a}{2d-3c}Rightarrowdfrac{b}{d}=dfrac{2b-3a}{2d-3c})

Áp dụng tính chất hoán vị ta được:

(dfrac{2d-3c}{d}=dfrac{2b-3a}{b})

Vậy (dfrac{2d-3c}{d}=dfrac{2b-3a}{b}left( ext{đ}pcm ight))

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

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

có \(1+\dfrac{2a}{3b}\ge2\sqrt{\dfrac{2a}{3b}}\)(BDT AM-GM)

\(=>1+\dfrac{2b}{3c}\ge2\sqrt{\dfrac{2b}{3c}}\)

\(=>1+\dfrac{2c}{3d}\ge2\sqrt{\dfrac{2c}{3d}}\)

\(=>1+\dfrac{2d}{3a}\ge2\sqrt{\dfrac{2d}{3a}}\)

\(=>S\ge16\sqrt{\dfrac{2a.2b.2c.2d}{3a.3b.3c.3d}}=16\sqrt{\dfrac{16abcd}{81abcd}}=16\sqrt{\dfrac{16}{81}}=\dfrac{64}{9}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)

a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)

b;c;d tương tự hết

b: a/b=c/d

nên 3a/3b=2c/2d

=>a/b=c/d=(3a+2c)/(3b+2d)

c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)

d: a/c=b/d

nên 5a/5c=2b/2d

=>a/c=b/d=(5a-2b)/(5c-2d)