lời giải đây nha cậu
a.\(\dfrac{2}{3}+x=\dfrac{5}{3}\)
x=\(\dfrac{5}{3}-\dfrac{2}{3}\)
x=1
b.\(3\dfrac{1}{2}-\dfrac{1}{2}.\left(x-\dfrac{3}{4}\right)=\dfrac{2}{3}\)
\(\dfrac{7}{2}-\dfrac{1}{2}.\left(x-\dfrac{3}{4}\right)=\dfrac{2}{3}\)
\(\dfrac{1}{2}.\left(x-\dfrac{3}{4}\right)=\dfrac{7}{2}-\dfrac{2}{3}\)
\(\dfrac{1}{2}.\left(x-\dfrac{3}{4}\right)=\dfrac{21}{6}+\dfrac{-4}{6}\)
\(\dfrac{1}{2}.\left(x-\dfrac{3}{4}\right)=\dfrac{17}{6}\)
x\(-\dfrac{3}{4}=\dfrac{17}{6}:\dfrac{1}{2}\)
\(x-\dfrac{3}{4}=\dfrac{17}{6}.2\)
\(x-\dfrac{3}{4}=\dfrac{17}{3}\)
x=\(\dfrac{17}{3}+\dfrac{3}{4}\)
x=\(\dfrac{68}{12}+\dfrac{9}{12}\)
x=\(\dfrac{77}{12}\)
c.3x=2y và x+y= 15
ta có: 3x=2y =>\(\dfrac{x}{2}=\dfrac{y}{3}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{15}{5}=3\)
vậy \(\dfrac{x}{2}=3=>x=2.3=6\)
\(\dfrac{y}{3}=3=>y=3.3=9\)
