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

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

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

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

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

\(\frac{1}{5}=\frac{b}{10}\Rightarrow b=2\)

mk không bít

mk mới lớp 7 thui

sorry nha

cảm ơn nhé

k mk nha

 k mk mk k lại

Bài này lp 8 mà cô giáo mk cho mk làm trong khi đó mk mới học lp 7

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.

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

\(\Leftrightarrow\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\)

\(\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\)

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

\(\Leftrightarrow40a+20-105+45a=0\)

\(\Leftrightarrow85a-85=0\)

\(\Leftrightarrow a=1\)

Thay a = 1 vào, ta được:

\(\dfrac{7-3}{20}=\dfrac{3b}{23+7}\)

\(\Leftrightarrow\dfrac{1}{5}=\dfrac{3b}{30}\)

\(\Leftrightarrow b=2\)

Vậy a = 1 ; b = 2.