\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)

vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)

Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)

Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)

      \(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)

Vậy...........

BAx = 35 nhé

Đặt \(\frac{x}{2}=\frac{y}{3}=k=>x=2k;y=3k\)

\(=>x.y=2k.3k=6k^2=54=>k^2=9=>k\in\left\{-3;3\right\}\)

+k=-3 , khi đó \(x=2.\left(-3\right)=-6;y=3.\left(-3\right)=-9\)

+k=3,khi đó \(x=2.3=6;y=3.3=9\)

Vậy.............
 

Đặt x/2 = y/3 = k

=> x= 2k

     y= 3k

=> xy = 54

2k * 3k = 54

6k^2 =54

k^2 = 9

k= +- 3

Th1: k= -3

=> x= 2k = 2* (-3) = -6

     y= 3k = 3* -3) = -9

Th2: k=3

=> x= 2k = 6

    y = 3k = 3*3 =9

Vậy x= +- 6

      y = +- 9

Ta có:

3x-1/2 = 0 

3x= 1/2

x= 1/6

và 1/2y + 3/5 =0

     1/2y = -3/5

         y= -6/5

Vậy x= 1/6 và y = -6/5

\(\left(3x-\frac{1}{2}\right)+\left(\frac{1}{2}y+\frac{3}{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{2}=0=\frac{1}{6}\\\frac{1}{2}y+\frac{3}{5}=0=\frac{6}{5}\end{cases}}\)

Vậy ......