lal+a

TH1 : = a + a = 2a

TH2 : = ( - a)  + a  = 0

             lal/a

TH1 : = a/a  =1

TH2 : = -a/a  = ( -1)

            lal-a

TH1 : = a - a = 0

TH2 : = (-a) - a 

           lal*a

TH1 : = a * a = a²

TH2 : = (-a) * a 

Ko chắc hok tốt

a)x=1
b)x=1
tick cho mình nha

a: \(\left|3x-1\right|+\left|2y-3\right|=0\)

=>\(\begin{cases}3x-1=0\\ 2y-3=0\end{cases}\Rightarrow\begin{cases}3x=1\\ 2y=3\end{cases}\Rightarrow\begin{cases}x=\frac13\\ y=\frac32\end{cases}\)

\(A=x^2+y^2=\left(\frac13\right)^2+\left(\frac32\right)^2=\frac19+\frac94=\frac{4+81}{36}=\frac{85}{36}\)

b: \(\left|x+2\right|+\left(2y-1\right)+\left|x+y+z-4\right|=0\)

=>\(\begin{cases}x+2=0\\ 2y-1=0\\ x+y+z-4=0\end{cases}\Rightarrow\begin{cases}x=-2\\ y=\frac12\\ z=4-x-y=4-\left(-2\right)-\frac12=4+2-\frac12=6-\frac12=\frac{11}{2}\end{cases}\)

\(B=x^2+y^2+z^2\)

\(=\left(-2\right)^2+\left(\frac12\right)^2+\left(\frac{11}{2}\right)^2=4+\frac14+\frac{121}{4}=4+\frac{122}{4}=\frac{122+16}{4}=\frac{138}{4}=\frac{69}{2}\)

a) A = 6X - 3 - X - 5

= (6X - X) - (-3 - 5)

= 5X - (-8)

= 5X + 8

b) B = 3X - 3 - 2X + 3

= (3X - 2X) - (-3 + 3)

= X +(3 - 3)

= X + 0 = X

c) C = @Alex Queeny

c) C = 2X - 6 + 4X - 1

= (2X + 4X) - (-6 - 1 )

= 6X - ( 6 + 1 )

= 6X - 7

Từ |12x-3| - |12x+3| = 0

Suy ra ta có 2 trường hợp:

(+) 12x - 3 = 0

\(\Rightarrow\) x = 1/4

(+) 12x + 3 = 0

\(\Rightarrow\) x = -1/4