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a) 2840 + [(999 - 9x) : 60 ] .24 = 3200

=> [(999-9x) : 60 ] . 24 = 2840 - 3200 = -360

=>  (999 - 9x) = \(\frac{-360}{24}=-15\)

=>  9x = 999 - ( - 15) = 999 + 15 = 1014

=>  x = \(\frac{1014}{9}=\frac{338}{3}\)

b) (3x - 48) . 6 = 3³.2² - 2³.3²

(3x - 48) . 6 = 27 . 4  -  8 . 9

(3x -48).6 = 36

(3x - 48 = 36 : 6 = 6

3x = 54

x = 54 : 3 =18

 t ick cho mik nha

1)

a)    2840 + [(999-9x) :60] . 24=3200

                 [(999-9x): 60] . 24=3200- 2840= 360

                (999-9x) : 60=    360 : 24

                (999-9x) : 60= 15

                999-9x        =15 . 60

                  999-9x       =900

                         9x=    999-900=9

                          x=9:9=1

b)   (3x - 48).6=    3³.2²-2³.3²

      (3x-48).6  =   27.4-8.9

       (3x - 48)6  =  108-72

        (3x - 48)6  = 36

       (3x-48)    =   36:6=6

        3x        = 6 + 48

        3x=54

          x=54:3=18

3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21

=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21

=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21

x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21

x . ( 1/2 - 1/14 ) = 1/21

x . 3/7 = 1/21 

x = 1/21 : 3/7

=> x = 1/9

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)

<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)

<=> \(x=\frac{1}{9}\)

\(=\left(1-x\right)\left(6a+2a^2\right)=2a\left(3+2a\right)\left(1-x\right)\\ 2,=\left(x-5\right)\left(x-3-2\right)=\left(x-5\right)^2\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

\(a,\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

vậy_____

Vì (x+1).(x-2)=0

suy ra 1 trong 2 so =0

tu giai not

a, vì (x+1).(x-2)=0 nên x+1=0  hoặc x-2=0

nếu x+1=0               nếu x-2=0

      x    =0-1                  x    =0+2

      x    =-1                    x    =2

Ta có : 1³+ 2³+...+100³ = (1+2+3+...+100)²  

Đặt A = 1+2+3+...+100

⇒A= \(\dfrac{\left(100+1\right).100}{2}\) ⇔A= 5050

⇒5050² = (x-1)² ⇔ x=5051

a) |2x-3|+x=21

|2x-3|=21-x

\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)

TH1: 2x-3=21-x

2x-x=21+3

x=24

TH2: 2x-3=-(21-x)

2x-3 = -21+x

2x-x=-21+3

x=-18

Vậy x \(\varepsilon\){-18;24}

4 . x³ + 15 = 47

4 . x³ = 47 - 15

4 . x³ = 32

x³ = 32 : 4

x³ = 8

=> x = 2

4 . 2^x - 3 = 125

4 . 2^x = 125 + 3

4 . 2^x = 128

2^x = 128 : 4

2^x = 32

=> x = 5

4 . x³ + 15 = 47

4 . x³ = 47 - 15

4 . x³ = 32

x³ = 32 : 4

x³ = 8

=> x = 2

(x + 2)(x - 1) - x(x + 3)

= x^2 - x + 2x - 2 - x^2 - 3x

= -2x - 2

giúp mình giải câu này vs

\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\) 

trả lời :

  (x+2)(x+1)-x(x+3)

=x²-x+2x-2-x²-3x

=-2x-2