a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

                                    \(2x=\frac{3}{5}-\frac{3}{5}\)

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

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

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

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

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

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

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

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a\(\left(x-3\right)^2-\left(x+2\right)^2-5\left(\frac{1}{5}-7\right)=-30\)

=>(x-3-x-2)(x-3+x+2)-x+35=-30

=>-5(2x-1)-x+35=-30

=>-10x+5-x+35=-30

=>-11x+40=-30

=>-11x=-70 =>x=70/11

d)\(\left(x+3\right)^2-\left(x+5\right)\left(x-5\right)=2\)

\(=>\left(x+3\right)^2-x^2+25=2\)

\(=>\left(z+3-z\right)\left(z+3+z\right)+25=2\)

\(=>3\left(2z+3\right)+25-2=0\)

\(=>6z+9+23=0\)

\(=>6x+32=0=>6x=-32=>x=-\frac{16}{3}\)

e)\(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

\(=>3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)

\(=>3x^2+12x+12+4x^2-4x+1-7x^2+63\)

\(=>8x+76=36=>8x=36-76=>x=-40\div8=-5\)

g)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(=>x^3-1-x\left(x^2-4\right)=5=>x^3-1-x^3+4x=5\)

\(=>4x-1=5=>4x=6=>x=\frac{3}{2}\)

ta có : x/3=y/9 => 2x/6=y/9

=> 2x/6=y/9=2x-y/6-9=12/-3=-4

+, 2x/6=-4 => x=-12

+, y/9=-4 => y=-36

a) 

Ta có : vì|1/2-1/3+x| lớn hơn hoặc bằng 0 

Còn -1/4-|y| bé hơn hoặc bằng 0

=> ko tồn tại x

b) 

Ta có: |x-y| lớn hơn hoặc bằng 0 và|y+9/25| lớn hơn hoặc bằng 0 mà:

| x-y|+ |y+9/25| =0 => |x-y| =0 và |y+9/25|=0

 Xét |y+9/25| có:

| y+9/25|=0 => y+9/25=0 => y=-9/25

Thay y = -9/25 vào |x-y| =0 => x=-9/25

 Vậy x=y=-9/25