b) \(x-\dfrac{4}{7}=\dfrac{2}{9}\)
\(x=\dfrac{2}{9}+\dfrac{4}{7}\)
\(x=\dfrac{50}{63}\)
\(x-\dfrac{4}{7}=\dfrac{2}{9}\)
\(x=\dfrac{4}{7}+\dfrac{2}{9}\)
\(x=\dfrac{50}{63}\)
x - 4/7 = 2/9
x = 2/9 + 4/7
x = 60/63
x = 20/21
Bài 1.\(x-\dfrac{4}{7}=\dfrac{2}{9}\)
\(x=\dfrac{2}{9}+\dfrac{4}{7}=\dfrac{50}{63}\)
Bài 2.
\(A=\dfrac{3n-4}{3n+2}=\dfrac{3n+2-6}{3n+2}=\dfrac{3n+2}{3n+2}-\dfrac{6}{3n+2}=1-\dfrac{6}{3n+2}\)
Để A nguyên thì \(\dfrac{6}{3n+2}\) nguyên hay \(3n+2\in U\left(6\right)=\left\{\pm1;\pm3;\pm6\right\}\)
3n+2=1 => n=-1/3
3n+2=-1 => n=-1
3n+2=3 => n=1/3
3n+2=-3 => n=-5/3
3n+2=6 => n=4/3
3n+2=-6 => n=-8/3
