a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

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

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

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

(2x-6)(3x-18)=0

=> 2x-6=0 hoặc 3x-18=0

=> x=3 hoặc x=6

47.(x-3)=47

=> x-3=1

=> x=4

a) X=3

b) X=73

c) X=4

   Câu 1 bạn gõ thiếu ngoặc nhé

a, [ 2x-6].[3x-18]=0

=> 2x-6=0    hoặc    3x-18=0

=>   2x =6             =>   3x =18

=>   x   =3             =>    x  =6

Vậy x thuộc{3 ,6}

kick mình nha

a)5. x +73.21=73.26

5. x +1533=1898

5.x             =1898-1533

5. x            =365

    x            =365:5

    x            =73

b)128-3.(x+4)=23

          3.(x+4)=128-23

          3.(x+4)=105

               x+4=105:3

                x+4=35

                x    =35-4

               x     =31

c)(x+1)+(x+3)+(x+4)+.....+(x+100)=5750

  suy ra (x+x+x+......+x)+(1+2+3+...+100)

suy ra 100.x + 5050=575 0

suy ra 100. x            =5750-5050

 suy ra 100. x           =700

                   x            =700:100

                   x            =7

Bạn tham khảo bài mình nhé .

a) 5 . x + 73 . 21 = 73 . 26

    5 . x + 1533 = 1898

    5 . x             = 1898 - 1533

    5 . x             = 365

         x             = 365 : 5

         x             = 73

Vậy x = 73

b) 128 - 3 ( x + 4 ) = 23

     3 ( x + 4 ) = 128 - 23

     3 ( x + 4 ) = 105

          x + 4    = 105 : 3

          x + 4    = 35

          x          = 35 - 4

          x          = 31

Vậy x = 31

c) ( x + 1 ) + ( x + 3 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750

\(\Rightarrow\) ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 )

\(\Rightarrow\) 100 . x + 5050 = 5750

\(\Rightarrow\) 100 . x             = 5750 - 5050

\(\Rightarrow\)100 . x              = 700

                  x              = 700 : 100

                  x              = 7

Vậy x = 7 .

cảm ơn nhé

thanks

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

x:6=24+6

x:6=30

x=30 x 6

x=180

23+x:2=37

x:2=37-23

x:2=14

x=14 x 2

x=28

214-136:x=197

136:x=214-197

136:x=17

x=136:17

x=8

(x-4):6:5=10

(x-4):6=10 x 5=50

(x-4)=50 x 6

x-4=300

x=300+4=304

(x+1)+(x+2)+...+(x+10)=755

10x + (1+2+...+10)=755

10x + (1+10) x [(10-1):1+1]:2=755

10x + 55 = 755

10x=755-55

10x=700

x=700:10=70

chúc bạn học tốt nha

a: \(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)

=>x=-1 hoặc x=1

b: \(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)

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

hay \(x\in\left\{-1;2;-2\right\}\)

c: \(x^3+x^2+4=0\)

\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)

\(\Leftrightarrow\left(x+2\right)\cdot\left(x^2-x+2\right)=0\)

=>x+2=0

hay x=-2

e: \(\Leftrightarrow x^4-2x^3-3x^3+6x^2-x^2+2x+3x-6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-3x^2-x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)=0\)

hay \(x\in\left\{2;3;-1;1\right\}\)

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

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