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

Ta có: \(\frac{1+2a}{15}=\frac{7-3a}{20}\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\Rightarrow20+40a=105-45a\)

           \(\Rightarrow85a=85\Rightarrow a=1\)

Thay a = 1 vào \(\frac{7-3a}{20}=\frac{3b}{23+7a}\), ta được:

           \(\frac{3b}{23+7}=\frac{7-3}{20}\Rightarrow\frac{3b}{30}=\frac{1}{5}\Rightarrow b=\frac{30}{3.5}=2\)

Vậy a = 1 , b = 2

\(\frac{1+2a}{15}=\frac{7-3a}{20}=\frac{3b}{23+7a}=\frac{3\left(1+2a\right)}{45}=\frac{2\left(7-3a\right)}{40}=\frac{17}{85}=\frac{1}{5}.\)

Vậy 1 + 2a = 3 => a = 1 

Thay vào: \(\frac{3b}{23+7\cdot1}=\frac{1}{5}\Rightarrow\frac{3b}{30}=\frac{1}{5}\Rightarrow b=2.\)

Vậy, a = 1 và b = 2.

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

Từ \(\frac{1+2a}{15}=\frac{7-3a}{20}\)\(\Rightarrow20+40a=105-45a\)

\(\Rightarrow a=1\)

Lại có \(\frac{1+2a}{15}=\frac{3b}{23+7a}\Rightarrow\frac{1}{5}=\frac{3b}{30}\Rightarrow b=2\)

Đáp số a=1,b=2

bn có thể làm rõ ra hơn đc ko bn Hoàng Thị Lan Hương

\(20\left(1+2a\right)=15\left(7-3a\right)\)

4(1+2a)=3(7-3a) => 8a +9a= 21-4 => 17a=17 => a=1

với a=1 ta có \(\frac{1+2a}{15}=\frac{3b}{23+7a}\)

\(\frac{1+2}{15}=\frac{3b}{23+7}\)1/5=b/10 => b=2

Vậy a= 1; b= 2

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

=\(\left(1+2a\right).\left(23+7a\right)=3b+15=\frac{7-3a}{20}\)

=\(\left[1+a\left(2+1\right)\right].\left[23+a\left(7+1\right)\right]=3b.15=\frac{7-3a}{20}\)

=[1+a3].[23+a8]=3b.15=7-3a/20

a[23+1(3+8+2)]=3b.15=7-3a/20

a.36=3b.15=7-3a/20

=>\(\frac{15}{a}=\frac{36}{3b}=\frac{7-3a}{20}=>\frac{15}{a}=12b=\frac{7-3a}{20}\)

con lai tu tinh nha

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

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

=> \(20\left(1+2a\right)=15\left(7-3a\right)\)

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

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

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

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

Vậy a =1 , b = 2