a-b=6

nên a=b+6

\(D=\dfrac{3\left(b+6\right)-6}{2\left(b+6\right)+b}-\dfrac{4b+6}{b+6+3b}\)

\(=\dfrac{3b+18-6}{2b+12+b}-1\)

\(=\dfrac{3b+12}{3b+12}-1=0\)

Ta có:\(\frac{3a-b}{2a+15}=\frac{3a-b}{2a+a-b}=\frac{3a-b}{3a-b}=1\)

          \(\frac{3b-a}{2b-15}=\frac{3b-a}{2b-\left(a-b\right)}=\frac{3b-a}{3b-a}=1\)  

=>P=1+1=2

Ta có a = 15 + b

=> \(\frac{3a-b}{2a+15}+\frac{3b-a}{2b-15}\) = \(\frac{3\left(15+b\right)-b}{2\left(15+b\right)+15}+\frac{3b-\left(15+b\right)}{2b-15}\)

= \(\frac{45+3b-b}{30+2b+15}+\frac{3b-15-b}{2b-15}\)

= \(\frac{45+2b}{45+2b}+\frac{2b-15}{2b-15}\)= 1 + 1 = 2

Ta có : \(\frac{3a+b+2a}{2a+c}=\frac{a+3b+c}{2b}=\frac{a+2b+2c}{b+c}\)

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

\(\Rightarrow\frac{a+b+c}{2a+c}+1=\frac{a+b+c}{2b}+1=\frac{a+b+c}{b+c}+1\)

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

\(\Rightarrow2a+c=2b=b+c\)

\(\Rightarrow\hept{\begin{cases}c=b\\a=\frac{1}{2}b\end{cases}}\)

Thay vào biểu thức trên , ta được :
\(P=\frac{\left(\frac{1}{2}b+b\right)\left(b+b\right)\left(b+\frac{1}{2}b\right)}{\frac{1}{2}b.b.b}\)

Vậy \(P=9\)

Trừ cả 3 đi 1 ta còn

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

Vói a+b+c=1 thì P=-1

Với a+b+c khác 0 thì

\(\Rightarrow2a+c=2b=b+c\Rightarrow2a=b=c\)

\(\Rightarrow P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\frac{3}{2}b2c3a}{abc}=9\)

Vậy............

THANK YOU

đặt b=3.a thì E=\(\frac{3a+9a}{4a-12a}=\frac{12a}{-8a}=-\frac{3}{2}\)