\(=\dfrac{5}{21}+\dfrac{16}{21}-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

a, 5/6 + 1/2 = 4/3

b, 7/8 + 1/4 = 9/8

c, 5/9 + 3/7 = 62/63

a,5/6+1/4=8/6 rút gọn =4/3   b,7/8+1/4=9/8   c,5/9+3/7=53/63

a,5/6+1/2=5/6+3/6=8/6

b,7/8+1/4=7/8+2/8=9/8

c5/9+3/7=35/63+27/63=62/63

Dùng công thức tính tổng 

\(1+2+3+...+x=11325\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=11325\)

\(\Leftrightarrow x\left(x+1\right)=22650\)

\(\Leftrightarrow x\left(x+1\right)=150.151\)

Nên x = 150

Vậy ,,,

\(1+2+3+4+...+x=11325\)

\(\Rightarrow\frac{x\left(x+1\right)}{2}=11325\)

\(\Rightarrow x\left(x+1\right)=11325\times2\)

\(\Rightarrow x\left(x+1\right)=22650\)

\(\Rightarrow150\times151=22650\)

\(\Rightarrow x=150\)

 (2^x-8)^3=(4^x+2^x+5)^3-(4^x+13)^3 
(2^x-8)^3=[(4^x+2^x+5)-(4^x+13)]*[(4^x... + (4^x+13)^2] 
(2^x-8)^3=(2^x-8)*[(4^x+2^x+5)^2+(4^x+... + (4^x+13)^2] 
2^x=8=>x=3 
hoặc (2^x-8)^2=(4^x+2^x+5)^2+(4^x+2^x+5)(4^x+... + (4^x+13)^2 
(4^x+2^x+5)^2 - (2^x-8)^2+(4^x+2^x+5)(4^x+13) + (4^x+13)^2=0 
[(4^x+2^x+5)-(2^x-8)]*[(4^x+2^x+5)+(2^... + (4^x+3)*[(4^x+2^x+5)+(4^x+13)]=0 
(4^x+13)*(4^x+2*2^x-3) + (4^x+3)*(2*4^x+2^x+18)=0 
(4^x+13)[(4^x+2*2^x-3) + (2*4^x+2^x+18)]=0 
4^x+13=0 (VN) 
hoặc 3*4^x + 3*2^x +15=0 
đặt t=2^x ( t>0) 
t^2 + t + 5=0 ptvn

a, 25/12

b,  25/72

a) = 3/4 + 4/3 = 25/12

b) = 3/8 - 16/9 = -101/72 * hình như hơi sai sai:"> *

a, ( 44 - x ) / 3 = ( x - 12 ) / 5

=> 5 ( 44 - x  ) = 3 ( x - 12 )

     220 - 5x     = 3x  - 36

     - 5x - 3x     = - 36 - 220

      - 8 x          = - 256

           x          = 32

b , ( 3 - x ) / 4 = ( 2x + 7 ) / 5

=> 5 ( 3 - x )   = 4 ( 2x + 7 )

     15 - 5x      = 8 x  + 28

     - 5 x - 8 x  = 28 - 15

        - 13 x     = 13

               x     = -1

a, \(\frac{\left(44-x\right)}{3}=\frac{\left(x-12\right)}{5}\)

 => (44 - x) . 5 = (x - 12) . 3

 => 44 - x . 5   = x - 12 .3

 => 44 - x . 5   = x - 36

 => x5 + x        = - 36 - 44

 => x5 + x        = - 80

=> x . (5 + 1)    = - 80

=> x . 6           = - 80

=> x                = - 80 : 6

=> x                = - 13,3

b, \(\frac{\left(3-x\right)}{4}=\frac{\left(2x+7\right)}{5}\)

=> (3 - x) . 5 = (2x + 7) . 4

=> 3 - x . 5   = 2x + 7 . 4

=> 3 - x . 5   = 2x + 28

=> -x . 5 + 2x = 28 - 3

=> -x . 5 + 2x = 25

=>  x . 5 + 2x = 25

=>  x . (5 + 2) = 25

=>  x . 7         = 25

=>  x              = 25 : 7

=>  x              = 3,57

a)(44-x).5=(x-12).3

44.5-x.5=x.3-12.3

220-x.5=x.3-36

220-x.5-x.3=-36

x.5-x.3=220+36

x.(5-3)=256

x.2=256

x=256:2

x=128

     \(x^2+2x+4⋮x+1\)

\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)

\(\Leftrightarrow3⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)

Ta có: \(x^2+2x+4\)

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

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

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

Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1

\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)