\(\frac{a}{3}=\frac{b}{8}=\frac{c}{5}\Rightarrow\frac{2a}{6}=\frac{3b}{24}=\frac{c}{5}=\frac{2a+3b-c}{6+24-5}=\frac{50}{25}=2\)

=> a/3 = 2 => a = 6

=> b/8 = 2 => b = 16

=> c/5 = 2 => c = 10

Đặt \(\frac{a}{3}=\frac{b}{8}=\frac{c}{5}=k\Rightarrow a=3k;b=8k;c=5k\)

=> \(2a+3b-c=6k+24k-5k=50\)

=> \(25k=50\Rightarrow k=2\)

=> \(\hept{\begin{cases}a=3\cdot2=6\\b=8\cdot2=16\\c=5\cdot2=10\end{cases}}\)

Ta có : \(\frac{a}{3}=\frac{b}{8}=\frac{c}{5}=\frac{2a}{6}=\frac{3b}{24}=\frac{2a+3b-c}{6+24-5}=\frac{50}{25}=2\)

Nên : \(\frac{a}{3}=2\Rightarrow a=6\)

         \(\frac{b}{8}=2\Rightarrow b=16\)

        \(\frac{c}{5}=2\Rightarrow x=10\)

Vậy ..........................

Lời giải:

$\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}$
$=\frac{5(3a-2b)}{25}=\frac{3(2c-5a)}{9}=\frac{2(5b-3c)}{4}$

$=\frac{5(3a-2b)+3(2c-5a)+2(5b-3c)}{25+9+4}=\frac{0}{25+9+4}=0$

$\Rightarrow 3a-2b=2c-5a=5b-3c=0$

$\Rightarrow 3a=2b; 2c=5a$

$\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{5}$

Áp dụng TCDTSBN:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{-50}{10}=-5$

$\Rightarrow a=(-5).2=-10; b=(-5).3=-15; c=(-5).5=-25$

Ta có :

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{15a-10b}{25}=\frac{6c-15a}{9}\)

\(=\frac{15a-10b+6c-15a}{25+9}=\frac{6c-10b}{34}=\frac{3c-5b}{17}=\frac{5b-3c}{2}\) = 0

=> a+b+c = 5a = - 50 => a = -10; b = -15 ; c = -25

Ta có : (3a-2b)/5 = (2c-5a)/3 <=> (15a-10b)/25 = (6c -15a)/9 = (15a-10b+6c-15a)/(25+9) = (3c-5b)/17 Do đó: (3c-5b)/17 = (5b-3c_

)/2 = 0. Nên 3a - 2b = 0 => b = 1,5a; 2c - 5a = 0 => c = 2,5a. Lúc đó : a+b+c= 5a = -50 => a = -10; b = -15, c= -25.

43x42=???

mong mọi người giúp đỡ