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

  \(\frac{12a-15b}{7}\)  = \(\frac{20c-12a}{9}\)  = \(\frac{15b-20c}{11}\) = \(\frac{12a-15b+20c-12a+15b-20b}{7+9+11}\) = \(\frac{0}{27}\) = 0

=> a = b = c

Mà a + b + c = 48

=> a = b = c = 48 : 3 = 16

Vậy a = b = c = 16.

\(\frac{12a-15b}{7}=\frac{20c-12a}{9}=\frac{15b-20c}{11}=\frac{12a-15b+20c-12a+15b-20c}{7+9+11}=0\)(tử bằng 0)

=> 12a - 15b = 20c - 12a = 15b - 20c => 12a = 15b = 20c

=>\(\frac{12a}{60}=\frac{15b}{60}=\frac{20c}{60}=\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{5+4+3}=\frac{48}{12}=4\)=> a = 4.5 = 20 ; b = 4.4 = 16 ; c = 4.3 = 12

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{12a-15b}{7}=\frac{20c-12a}{9}=\frac{15b-20c}{11}=\frac{12a-15b+20c-12a+15b-20c}{7+9+11}=0\)

\(\frac{12a-15b}{7}=0\Rightarrow12a=15b\Rightarrow\frac{a}{15}=\frac{b}{12}\Rightarrow\frac{a}{5}=\frac{b}{4}\)(1)

\(\frac{20c-12a}{9}=0\Rightarrow20c=15a\Rightarrow\frac{a}{20}=\frac{c}{12}\Rightarrow\frac{a}{5}=\frac{c}{3}\)(2)

\(\frac{15b-20c}{11}=0\Rightarrow15b=20c\Rightarrow\frac{b}{20}=\frac{c}{15}\Rightarrow\frac{b}{4}=\frac{c}{3}\)(3)

từ (1),(2),(3) => \(\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{5+4+3}=\frac{48}{12}=4\)(t/c dãy tỉ số bằng nhau)

\(\frac{a}{5}=4\Rightarrow a=20,\frac{b}{4}=4\Rightarrow b=16,\frac{c}{3}=4\Rightarrow c=12\)

Vậy a=20, b=16, c=12

Áp dụng tc của dãy tỉ số bằng nhau :

\(\frac{12a-15b}{7}=\frac{20c-12a}{9}=\frac{15b-20c}{11}=\frac{12a-15b+20c-12a+15b-20c}{7+9+11}=\frac{0}{27}=0\)

\(=>\hept{\begin{cases}12a-15b=0=>12a=15b=>\frac{a}{5}=\frac{b}{4}\\20c-12a=0=>20c=12a=>\frac{c}{3}=\frac{a}{5}\\15b-20c=0=>15b=20c=>\frac{c}{3}=\frac{b}{4}\end{cases}=>\frac{a}{5}=\frac{b}{4}=\frac{c}{3}}\)

Đặt \(\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=k=>\hept{\begin{cases}a=5k\\b=4k\\c=3k\end{cases}}\)

Thay vào : \(a+b+c=5k+4k+3k=12k=48=>k=4\)

\(=>\hept{\begin{cases}a=5k=5.4=20\\b=4k=4.4=16\\c=3k=3.4=12\end{cases}}\)

Vậy...

a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)

\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)

\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)

b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)

\(=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x\left(x+3\right)}\)

c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)

\(=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}\)

\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)

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

d) \(\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)\)

\(=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}\)

\(=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}\)

\(=x+\dfrac{2-x^2}{x\left(x-1\right)}\)

a: a=75; b=135