dấu ''='' k xảy ra nên chỉ cm đc > hơn thôi nhé

\(a^2+9b^2+c^2+9,5>2a+12b+4c\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(9b^2-12b+4\right)+\left(c^2-4c+4\right)>0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(3b-2\right)^2+\left(c-2\right)^2+0,5>0\) --> luôn đúng

-->đpcm

_a2_-2a+1+4b2_-12b+9+3c2_-6c+3+1>0

⇔(a−1)2+(2b−3)2+3(c−1)2+1>0 (luôn đúng)

$\Rightarrow$ BĐT ban đầu đúng

Ta có:

\(a^2+9b^2+c^2+\dfrac{19}{2}-2a-12b-4c=a^2-2a+1+9b^2-12b+4+c^2-4c+4+\dfrac{1}{2}=\left(a-1\right)^2+\left(3b-2\right)^2+\left(c-2\right)^2+\dfrac{1}{2}>0\left(1\right)\)Vì (1) luôn đúng nên \(a^2+9b^2+c^2+\dfrac{19}{2}>2a+12b+4c\)

\(\Leftrightarrow a^2-2a+1+4b^2-12b+9+3c^2-6c+3+1>0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(2b-3\right)^2+3\left(c-1\right)^2+1>0\) (luôn đúng)

\(\Rightarrow\) BĐT ban đầu đúng

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

a: \(\frac{2a+5b}{3a-4b}=\frac{2\cdot bk+5b}{3\cdot bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\frac{2c+5d}{3c-4d}=\frac{2\cdot dk+5d}{3\cdot dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

Do đó: \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

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

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

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

d: \(\frac{4a+9b}{4a-7b}=\frac{4\cdot bk+9b}{4\cdot bk-7b}=\frac{b\left(4k+9\right)}{b\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

\(\frac{4c+9d}{4c-7d}=\frac{4\cdot dk+9d}{4\cdot dk-7d}=\frac{d\left(4k+9\right)}{d\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

Do đó: \(\frac{4a+9b}{4a-7b}=\frac{4c+9d}{4c-7d}\)