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

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

3a = 2b => a/2 = b/3  => \(\frac{a}{10}=\frac{b}{15}\) (1)

7b = 5c => b/5 = c/7  => \(\frac{b}{15}=\frac{c}{21}\) (2)

Từ (1) và (2) => \(\frac{a}{10}=\frac{b}{15}=\frac{c}{21}\)

Áp dụng dãy tỉ số bằng nhau  ta có :

                    \(\frac{a}{10}=\frac{b}{15}=\frac{c}{21}=\frac{a-b+c}{10-15+21}=\frac{32}{16}=2\)

=> a = 2.10 = 20 

=> b = 2.15 = 30 

=> c = 2.21 = 42 

Tich đúng cho mình nha bạn 

\(2a=2b\Rightarrow\frac{a}{2}=\frac{b}{2}\Rightarrow\frac{a}{2}.\frac{1}{7}=\frac{b}{2}.\frac{1}{7}\Rightarrow\frac{a}{14}=\frac{b}{14}\)
\(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{7}.\frac{1}{2}=\frac{c}{5}.\frac{1}{2}\Rightarrow\frac{b}{14}=\frac{c}{10}\)
(Ngoặc '}' 2 điều trên lại)
\(\Rightarrow\frac{a}{14}=\frac{b}{14}=\frac{c}{10}\)(1)
Từ (1) \(\Rightarrow\frac{3a}{3.14}=\frac{7b}{7.14}=\frac{5c}{5.10}=\frac{3a}{42}=\frac{7b}{98}=\frac{5c}{50}\)
Áp dụng tính chất DTSBN:
\(\frac{a}{14}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{42}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{42-98+50}=\frac{-30}{-6}=5\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{14}=5\Rightarrow a=5.14=70\\\frac{b}{14}=5\Rightarrow a=5.14=70\\\frac{c}{10}=5\Rightarrow c=5.10=50\end{cases}}\)
Vậy a = 70, b = 70, c = 50

Cảm ơn nha!

3a=2b => a/2 = b/3  => a/14 = b/21

5b = 7c => b/7 = c/5 => b/21 = c/15

=> a/14 = b/21= c/15

= \(\frac{3a}{42}=\frac{7b}{147}=\frac{5c}{75}=\frac{3a+5c-7b}{42+75-147}=\frac{60}{-30}=-2\)

=> a = -2.14 = -28

b = -2.21 = -42

c= -2 . 15= -30

kết quả là 1,2

a, Ta có: \(\frac{a}{3}=\frac{b}{2};\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\frac{a}{21}=\frac{b}{14}=\frac{c}{15}=\frac{3a-7b+5c}{63-98+75}=\frac{30}{40}=\frac{3}{4}\)

\(a=\frac{63}{4};b=\frac{42}{4};c=\frac{45}{4}\)

b, Ta có : \(7a=9b=21c\Rightarrow\frac{7a}{63}=\frac{9b}{63}=\frac{21c}{63}\Rightarrow\frac{a}{9}=\frac{b}{7}=\frac{c}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\frac{a}{9}=\frac{b}{7}=\frac{c}{3}=\frac{a-b+c}{9-7+3}=-\frac{15}{5}=-3\Rightarrow a=-27;b=-21;c=-9\)