Lời giải:

a)\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow4a=3b\)

Và \(4a.5=3b.5\Leftrightarrow20a=15b\Leftrightarrow\dfrac{20a}{3}=5b\)

Khi đó:

\(A=\dfrac{2a-5b}{a-3b}=\dfrac{2a-\dfrac{20}{3}a}{a-4a}=\dfrac{-\dfrac{14}{3}a}{-3a}=\dfrac{-14}{\dfrac{3}{-3}}=14\)

b) Ta có:

\(a-b=7\Leftrightarrow b=a-7\)

\(B=\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{3a-\left(a-7\right)}{2a+7}+\dfrac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)

\(B=\dfrac{3a-a+7}{2a+7}+\dfrac{3a-21-a}{2a-14-7}\)

\(B=\dfrac{2a+7}{2a+7}+\dfrac{2a-21}{2a-21}=1+1=2\)

***** Ta có       \(A=\frac{2a-5b}{a-3b}\)Mà \(\frac{a}{b}=\frac{6}{8}\Leftrightarrow b=\frac{8a}{6}=\frac{4}{3}a\)Thay b vào biểu thức A , ta có : \(\frac{2a-5.\frac{4}{3}a}{a-3.\frac{4}{3}a}=\frac{a\left(2-5.\frac{4}{3}\right)}{a\left(1-3.\frac{4}{3}\right)}=\frac{-14}{3}:\left(-3\right)=\frac{14}{9}\)Vậy \(A=\frac{14}{9}\)

***** Ta có \(B=\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}\)MÀ a-b=7 => a = b+7  . Thay a = b+7 vào biểu thức B , ta có \(\frac{3.\left(7+b\right)-b}{2\left(7+b\right)+7}+\frac{3b-\left(7+b\right)}{2b-7}=\frac{21+3b-b}{14+2b+7}+\frac{3b-7-b}{2b-7}\)=>>>>> \(\frac{21+2b}{21+2b}+\frac{2b-7}{2b-7}=1+1=2\)(k mình nha )

\(a-b=7\Leftrightarrow b=a-7\)

\(\Rightarrow P=\frac{3a-\left(a-7\right)}{2a-7}+\frac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)

\(=\frac{3a-a+7}{2a-7}+\frac{3a-21-a}{2a-14-7}\)

\(=\frac{2a+7}{2a-7}+\frac{2a-21}{2a-21}\)

\(=\frac{2a+7}{2a-7}+1=\frac{2a+7+2a-7}{2a-7}=\frac{4a}{2a-7}\)

Ta có :

\(\frac{a}{b}=\frac{3}{4}\)\(\Rightarrow\)\(a=3k;b=4k\)\(\left(k\in\right)ℤ\)

Suy ra :
\(\frac{2a-5b}{a-3b}=\frac{6k-20k}{3k-12k}=\frac{k\left(6-20\right)}{k\left(3-12\right)}=\frac{-14}{-9}=\frac{14}{9}\)

Ta có  a - b = 7 => a = 7 + b 

Thay a = 7+b vào C có : 

\(C=\frac{3\left(7+b\right)-b}{2\left(7+b\right)+7}+\frac{3b-7-b}{2b-7}\)

\(C=\frac{21+3b-b}{14+2b+7}+\frac{2b-7}{2b-7}\)

\(C=\frac{21+2b}{21+2b}+1=1+1=2\)

Vậy \(C=2\)

Ta có:\(a-b=7\Leftrightarrow7=a-b\)

           Thay \(7=a-b\)vào biểu thức,ta được:

                \(\frac{3a-b}{2a+7}+\frac{3a-b}{2b-7}=\frac{3a-b}{2a+a-b}+\frac{3a-b}{2b-a+b}\)

                                                       \(=\frac{3a-b}{3a-b}+\frac{3b-a}{3b-a}\)

                                                       \(=1+1\)

                                                       \(=2\)

                                        Vậy giá trị của biểu thức C=2

a)Thay \(x=\dfrac{-2}{3}\) vào\(x^3-6x^2-9x-3\):

\(\left(\dfrac{-2}{3}\right)^3-6\left(\dfrac{-2}{3}\right)^2+9.\dfrac{2}{3}-3\)

\(=\dfrac{-8}{27}-\dfrac{8}{3}+6-3\)

\(=\dfrac{-8-72}{27}+3=\dfrac{-80}{27}+3=\dfrac{1}{27}\)

b) Ta có: \(\dfrac{a}{b}=\dfrac{3}{4}\Rightarrow a=3k;b=4k\)

\(\Rightarrow\dfrac{2a-5b}{a-3b}=\dfrac{6k-20k}{3k-12k}=\dfrac{-14k}{-9k}=\dfrac{14}{9}\)

c) Có: a-b=7\(\Rightarrow a=b+7\)

Thay vào \(\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{2b+21}{2b+21}+\dfrac{2b-7}{2b-7}\)

\(=1+1=2\)

Ta có: \(a-b=7\)

\(\Rightarrow b-a=-7\)

\(B=\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}\)

\(B=\frac{2a+\left(a-b\right)}{2a+7}+\frac{2b+\left(b-a\right)}{2b-7}\)

\(B=\frac{2a+7}{2a+7}+\frac{2b-7}{2b-7}\)

\(B=1+1\)

\(B=2\)

Vậy \(B=2\)

Tham khảo nhé~

    \(B=\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}\)

       \(=\frac{2a+\left(a-b\right)}{2a+7}+\frac{2b-\left(a-b\right)}{2b-7}\)

       \(=\frac{2a+7}{2a+7}+\frac{2b-7}{2b-7}\) (vì a - b = 7)

       \(=1+1=2\)

kết quả là 1+1=2

nhớ nhé