cách khác:

\(B=\frac{3a-2b}{2a+5}+\frac{3b-a}{b-5}\)

\(=\frac{3a-2b}{2a+a-2b}+\frac{3b-a}{b-a+2b}\)  (thay 5 = a - 2b)

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

\(=1+1=2\)

Biết a - 2b = 5 tính giá trị biểu thức:

\(B=\frac{3a-2b}{2a+5}+\frac{3b-a}{b-5}\)

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

\(=\frac{2a+5}{2a+5}+\frac{b-5}{b-5}\)

\(=1+1=2\)

Vậy B = 2

a-2b=5 => a=2b+5

Thay a=2b+5 vào B thì : 

B = 6b+15-2b/4b+10+5 + 3b-2b-5/b-5

   = 4b+15/4b+15 + b-5/b-5 = 1+1 = 2

Tk mk nha

Ta có : a - 2b = 5 \(\Rightarrow\)2b = a - 5

          a - 2b = 5 \(\Rightarrow\)a = 2b + 5

Thay vào , ta được :

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

\(B=\frac{3a-a+5}{2a+5}+\frac{3b-2b-5}{b-5}\)

\(B=\frac{2a+5}{2a+5}+\frac{b-5}{b-5}\)

\(B=1+1=2\)

\(B=\frac{3a-2b}{2a+5}+\frac{3b-a}{b-5}\)

\(B=\frac{2a+\left(a-2b\right)}{2a+5}+\frac{b-\left(a-2b\right)}{b-5}\)

\(B=\frac{2a+5}{2a+5}+\frac{b-5}{b-5}\)

\(B=1+1=2\)

Vậy 

Từ a-2b=5  =>  a = 2b+5 

Thay 2b + 5 vào a, ta có biểu thức  :

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

\(=\frac{6b+15-2b}{4b+10+5}+\frac{3b-2b-5}{b-5}=\frac{4b+15}{4b+15}+\frac{b-5}{b-5}=1+1=2\)

thay a-2b vào biểu thức cần tính

thay a-2b vào biểu thức cần tính

Bài này dễ mà bạn

dễ thì bn giải hộ mk đi,nói đc lm đc nhỉ

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

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

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

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

Từ \(a-2b=5\Rightarrow a=5+2b\) thay vào P ta có:

\(P=\frac{3\left(2b+5\right)-2b}{2\left(2b+5\right)+5}+\frac{3b-\left(2b+5\right)}{b-5}\)\(=\frac{6b+15-2b}{4b+10+5}+\frac{3b-2b+5}{b-5}\)

\(=\frac{4b+15}{4b+15}+\frac{b-5}{b-5}=1+1=2\)