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

***** 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 )

Ta có :

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

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

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

mà 2a-b=7 hay b-2a=-7 nên ta có

\(P=1+\frac{7-7}{3a+7}-1-\frac{-7+7}{2b-7}=1+0-1-0=0\)

Ta có: \(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\Leftrightarrow85a=85\Leftrightarrow a=1\)

Ta có: \(\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\Rightarrow\dfrac{7-3.1}{20}=\dfrac{3b}{23+7.1}\Rightarrow\dfrac{4}{20}=\dfrac{3b}{30}\Rightarrow\dfrac{1}{5}=\dfrac{b}{10}\Rightarrow b=2\) Vậy a=1;b=2

TL :

Ta có : \(\frac{1+2a}{15}=\frac{7-3a}{20}=\frac{3b}{23+7a}\)

Vì \(\frac{1+2a}{15}=\frac{7-3a}{20}\)

\(\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\)

\(\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\)

\(\Leftrightarrow95a=95\Rightarrow a=1\)

Thay a = 1 vào phương trình  \(\frac{7-3a}{20}=\frac{3b}{23+7a}\); ta có : \(\frac{7-3.1}{20}=\frac{3b}{23+7.1}\)

\(\Leftrightarrow\frac{4}{20}=\frac{3b}{30}\Leftrightarrow\frac{1}{5}=\frac{b}{10}\Leftrightarrow5b=10\Rightarrow b=2\)

Vậy a = 1 ; b = 2

Có:

\(\frac{1+2a}{15}=\frac{7-3a}{20}\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\Rightarrow a=1\)

Thay a=1 vào\(\frac{1+2a}{15}=\frac{3b}{23+7a}=\frac{1}{5}=\frac{b}{10}\Rightarrow b=2\)