\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\Leftrightarrow a=b=c\)

\(A=\frac{a}{b+c}+\frac{a+b}{c}=\frac{5}{2}\)

\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\).

\(\frac{5a+7b}{5c+7d}=\frac{5bk+7b}{5dk+7d}=\frac{b\left(5k+7\right)}{d\left(k+7\right)}=\frac{b}{d}\)

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

Suy ra đpcm.

Đk: x \(\ge\)0

A = \(\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)

Do \(\sqrt{x}\ge\)0 => \(\sqrt{x}+1\ge1\)

=> \(\frac{2}{\sqrt{x}+1}\le2\) => \(1+\frac{2}{\sqrt{x}+1}\le2+1=3\)

=> A \(\le\)3

Dấu "=" xảy ra <=> x = 0

Vậy MaxA = 3 khi x = 0

Ta có : \(\hept{\begin{cases}2x=3y\\5y=7z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{3}=\frac{y}{2}\\\frac{y}{7}=\frac{z}{5}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{21}=\frac{y}{14}\\\frac{y}{14}=\frac{z}{10}\end{cases}}\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Đặt \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=k\Rightarrow\hept{\begin{cases}x=21k\\y=14k\\z=10k\end{cases}}\)

Khi đó 3x - 7y + 5z  = 30

<=> 3.21k - 7.14k + 5.10k = 30

=> 63k - 98k + 50k = 30

=> 15k = 30

=> k = 2

=> x = 42 ; y = 28 ; z = 20

Vậy  x = 42 ; y = 28 ; z = 20 là giá trị cần tìm

Ta có : \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)

\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\)

Ta lại có : \(\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)(*)

\(\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)(**)

Từ (*) ; (**) =)) \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{20}=\frac{3}{2}\)

\(x=\frac{63}{2};y=\frac{42}{2}=21;z=\frac{30}{2}\)