Vì: \(\left|x+\frac{13}{14}\right|\ge0\forall x;\left|y-\frac{8}{27}\right|\ge0\forall y\)

mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{8}{27}\right|=0\)

\(\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{8}{27}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{8}{27}\end{cases}}\)

Vậy \(x=-\frac{3}{4};y=\frac{8}{27}\)

Chữ A ngược nghĩa là gì vậy bạn

\(\forall\)có nghĩa là "với mọi..." nha bạn

b, 

ta có: x-12/3 + y+8/23 + z+190/27 luôn lớn hơn 0 nên không thể nhỏ hơn 0

Để: |x-12/3| + |y+8/23| + |z+190/27| > 0

=> (+) x-12/3 = 0

=> x= 12/3

(+) y+8/23 = 0

=> y = -8/23

(+) z+190/27 = 0

=> z = -190/27

Vậy x = 12/3; y = -8/23; z = -190/27

k giúp mình

làm ơn

câu a sai đề thì phải, bạn chữa lại rồi mình làm

sai đề thật rồi đấy

\(1,x^2+2xy+x+2y\)

\(=\left(x^2+2xy\right)+\left(x+2y\right)\)

\(=x\left(x+2y\right)+\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+1\right)\)

\(2,x^2-10x+25\)

\(=x^2-2.x.5+5^2\)

\(=\left(x-5\right)^2\)

Đợi mk chút ,mk có việc bận ,tối mk làm tiếp nha bn

\(3,x^3+3x^2+3x+1\)

\(=\left(x^3+1\right)+\left(3x^2+3x\right)\)

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

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

\(=\left(x+1\right)\left(x^2+2x+1\right)\)

\(=\left(x+1\right)\left(x+1\right)^2\)

\(=\left(x+1\right)^3\)

\(4,x^3-8\)

\(=x^3-2^3\)

\(=\left(x-2\right)\left(x^2+2x+4\right)\)

\(5,x^3+27\)

\(=x^3+3^3\)

\(=\left(x+3\right)\left(x^2-3x+9\right)\)

\(6,x^3-\dfrac{1}{8}\)

\(=x^3-\left(\dfrac{1}{2}\right)^3\)

\(=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)

\(7,x^3-x+y^3-y\)

\(=\left(x^3+y^3\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

\(8,4x^2-1\)

\(=\left(2x\right)^2-1^2\)

\(=\left(2x-1\right)\left(2x+1\right)\)

\(9,49x^2-9\)

\(=\left(7x\right)^2-3^2\)

\(=\left(7x-3\right)\left(7x+3\right)\)

a. Tính đúng y = 40

b. Tính đúng  y = 4 2 3

a) (-12) . (x - 14) + 7 . (3 - x) = 12.15

=> -12x + 168 + 21 - 7x = 180

=> -19x = -9

=> x = \(\frac{9}{19}\)

b) (2x - 8) . ( y - 2) = 0

=> \(\left\{\begin{matrix}2x-8=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=4\\y=2\end{matrix}\right.\)

c) x^2 - 75 = (-50)

=> \(x^2\)=25

=> \(\left[\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

d) (2x - 5)^2 - 1 = 0

=> ( 2x - 5 ) ^2 = 1

=> \(\left[\begin{matrix}2x-5=1\\2x-5=-1\end{matrix}\right.\)

=>\(\left[\begin{matrix}x=3\\x=2\end{matrix}\right.\)

e) (3y + 6)^3 + 27 = 0

=> \(\left(3y+6\right)^3=-27\)

=> 3y + 6 = -3

=> y = -3

1a) 25 .8 - 12.5 + 272 :17 - 8

= 200 -  60 + 16 - 8

= 140+16-8

= 156 - 8

= 148

b) 125: 25 +14 - 142 : 71

= 5 + 14 - 2 

= 19 - 2 = 17

c) 13. 17 - 256: 16 + 14 : 7 - 1

= 221 - 16 + 2 - 1 

= 205 + 2 - 1 

= 207 - 1 = 206

d) 15 . 24 - 14 . 5 (145 : 5 - 27)

= 360 - 70 (29- 27)

= 360 - 70 . 2

= 360 - 140

= 220

2a) 37 + x = 55

               x= 55 - 37

               x = 18

b) x- 31 = 24

     x       = 24 + 31

     x        = 55

c) 21 (x-11) = 21

        x - 11   = 21 : 21

          x        = 1 +11

           x       = 12

d ) 9. (x - 29)= 0

           x - 29 = 0 :9

             x       = 29

e) (x - 1954) . 5 = 50

         x - 1954   = 50 : 5

          x             = 1964

f ) 30 . 60 - x = 30

        1800 - x   = 30

             x        = 1800 - 30

             x        = 1770

Viết mỏi tay quá huhuh

Cảm ơn nghen

thanks bn nhé

\(2x\left(2y-14\right)-8\left(y-7\right)=0\)

=>\(2x\cdot2\cdot\left(y-7\right)-8\left(y-7\right)=0\)

=>\(4x\left(y-7\right)-8\left(y-7\right)=0\)

=>\(\left(4x-8\right)\left(y-7\right)=0\)

=>\(\left\{{}\begin{matrix}4x-8=0\\y-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=7\end{matrix}\right.\)