\(A=\frac{3x+2}{5x+3}\)
Gọi d là ƯC(3x+2 ; 5x+3)
\(\Rightarrow\hept{\begin{cases}3x+2⋮d\\5x+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}5\left(3x+2\right)⋮d\\3\left(5x+3\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}15x+10⋮d\\15x+9⋮d\end{cases}}\)
\(\Rightarrow\left(15x+10\right)-\left(15x+9\right)⋮d\)
\(\Rightarrow15x+10-15x-9⋮d\)
\(\Rightarrow1⋮d\Leftrightarrow d=1\)
=> ƯCLN( 3x+2 ; 5x+3 ) = 1
=> \(A=\frac{3x+2}{5x+3}\)tối giản ( đpcm )
\(B=\frac{2x+3}{4x+8}\)
Gọi d là ƯC( 2x+3;4x+8 }
\(\Rightarrow\hept{\begin{cases}2x+3⋮d\\4x+8⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2\left(2x+3\right)⋮d\\4x+8⋮d\end{cases}}\Rightarrow\hept{\begin{cases}4x+6⋮d\\4x+8⋮d\end{cases}}\)
\(\Rightarrow\left(4x+8\right)-\left(4x+6\right)⋮d\)
\(\Rightarrow4x+8-4x-6⋮d\)
\(\Rightarrow2⋮d\Leftrightarrow d=\left\{1;2\right\}\)
Với d = 1 => \(2x+3⋮d\)
Với d = 2 => \(2x+3⋮̸d\)vì \(3⋮̸2\)
=> d = 1
=> ƯCLN( 2x+3 ; 4x+8 ) = 1
=> \(B=\frac{2x+3}{4x+8}\)tối giản ( đpcm )
