BT1 : Tính giá trị của biểu thức ;

Thay 7 = a -b vào biểu thức B ,có :

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

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

\(=1+1\)

= 2

Vậy giá trị của biểu thức B là 2 với a- b=7

Ta có :

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

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

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

\(=1+1=2\)

a-b=7 nên a=b+7

\(P=\dfrac{3\left(b+7\right)-b}{2\left(b+7\right)+7}+\dfrac{3b-b-7}{2b-7}=1+1=2\)

\(\dfrac{5a-b}{3a+7}\)-\(\dfrac{3b-2a}{2b-7}\)

=\(\dfrac{5a-b}{3a+2a-b}\)-\(\dfrac{3b-2a}{2b-\left(2a-b\right)}\)

=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{2b-2a+b}\) (vì 2a-b=7)

=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{3b-2a}\)

=1-1

=0

A=\(\dfrac{5}{9}\)

\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)

A=5/9

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

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

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

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

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