\(\Leftrightarrow\dfrac{1}{2}x+\dfrac{2}{3}x-x=-4\Leftrightarrow\dfrac{3x+4x-6x}{6}=-\dfrac{24}{6}\)

\(\Rightarrow x=-24\)

b) \(x.\left(\dfrac{1}{2}+\dfrac{2}{3}-1\right)=-4\)

\(x.\dfrac{-1}{6}=-4\)

\(x=-4:\dfrac{-1}{6}\)

\(x=-24\)

Ta có: \(0.5x+\dfrac{2}{3}x-x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{6}=-4\)

hay x=-24

Lời giải:

$50=4\times y+2$

$50-2=4\times y$

$48=4\times y$

$48:4=y$

$12=y$

y=12

y= 12

\(35-5\left(x-1\right)=10\\ \Leftrightarrow35-5x+5=10\\ \Rightarrow40-5x=10\)

\(\Rightarrow-5x=10-40\\ \Rightarrow-5x=-30\\ \Rightarrow x=\dfrac{-30}{-5}=6\)

c) 

\(24\left(x-16\right)=12^2\)

\(\Rightarrow24x-384=144\\ \Rightarrow24x=144+384\\ \Rightarrow24x=528\\ \Rightarrow x=\dfrac{528}{24}=22\)

d) 

\(\left(x^2-10\right)\div5=3\\ \Rightarrow\left(x^2-10\right)=3\times5\\ \Rightarrow x^2-10=15\)

\(\Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\Rightarrow x=5\)

Ơ, bn ơi @Kenny sai sai ở đâu thì phải.

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

Bài 2:

Với x,y,z,t là số tự nhiên khác 0

Có \(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)

\(\dfrac{y}{x+y+z+t}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)

\(\dfrac{z}{x+y+z+t}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)

\(\dfrac{t}{x+y+z+t}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)

Cộng vế với vế \(\Rightarrow1< M< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}=2\)

=> M không là số tự nhiên.

Bài 1:

Ta có:

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\) 

\(B=\left(1+\dfrac{2007}{2}\right)+\left(1+\dfrac{2006}{3}\right)+...+\left(1+\dfrac{2}{2007}\right)+\left(1+\dfrac{1}{2008}\right)+1\) 

\(B=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}\) 

\(B=2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}=2009\)

sai rồi kìa \(\frac{A}{B}\)chớ không phải \(\frac{B}{A}\)

bằng \(\frac{1}{2009}\)mới dúng

Ta có : 

\(\left|3x+18\right|\ge0\) và \(\left|4x-28\right|\ge0\) \(\Rightarrow\) \(\left|3x+18\right|+\left|4y-28\right|\ge0\)

Mà \(\left|3x+18\right|+\left|4y-28\right|\le0\) ( đề bài cho )

\(\Rightarrow\)\(\left|3x+18\right|+\left|4y-28\right|=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+18=0\\4y-28=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-18\\4y=28\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\y=7\end{cases}}}\)

Vậy \(x=-6\) và \(y=7\)

Ta có \(\left|3x+18\right|+\left|4y-28\right|\le0\)

Mà \(\left|3x+18\right|\ge0\forall x;\left|4y-28\right|\ge0\forall y\)

=> |3x+18|+|4y-28|=0

=> 3x+18=4y-28=0

• 3x+18=0 <=> 3x=-18 <=> x=-6

• 4y-28=0 <=> 4y=28 <=> y=7

Vậy ...

thank you very much

24 : ( 3x -2 ) = -3

       ( 3x - 2 ) = 24 : -3 

       ( 3x -2 ) = -8

       3x          = -8 + 2

       3x          = -6

         x          = -6 : 3

         x           = -2

3x-2 = 24: (-3)

3x-2 = -8

3x = (-8) +2

3x = -6

x = -6:3

x = -2

c) 24 : (3x - 2) = -3

            3x - 2  = 24 : (-3)
            3x - 2  = -8

            3x      = (-8) + 2

            3x      = -6

              x      = (-6) : 3

              x      = -2

\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)

a: =>2x-5=-11

=>2x=-6

hay x=-3

b: =>2(4-3x)=14

=>4-3x=7

=>3x=-3

hay x=-1

c: =>3x-2=-8

=>3x=-6

hay x=-2

a) 2x-5+17=6

2x+12=6

2x=12-6

2x=6

x=6:2

x-3

b) 10-8+6x=-4

2+6x=-4

6x=-4-2

6x=-6

x=(-6):6

x=-1

c) 3x-2=24:(-3)

3x-2=-8

3x=-8+2

3x=-6

x=-6:3

x=-2

d) 5-2x=-5

-2x=-5-5

-2x=-10

x=-10:(-2)

x=5

1) 3x = 45 + 15 = 60

x = 60 : 3 = 20

2) 5x = 50 - 35 = 15

x = 15 : 5 = 3

3) (2x - 5) + 17 = 6

2x - 5 = 6 - 17

2x - 5 = -11

2x = -11 + 5

2x = -6

x = -6 : 2 = -3

4) 10 - 2(4 -3x) = -4

2(4 - 3x) = -4 - 10 = -14

4 - 3x = -14 : 2 = -7

3x = -7 - 4 = -18

x = -18 : 3 = -6