a) 6x = 19 + 15 = 34

=> x = 34 : 6 = 17/3

b) 3x + 2 = 8 

=> 3x = 8 - 2 = 6

=> x = 6 : 3 = 2

c) 12x - 33 = 135

=> 12x = 135 + 33 = 168

=> x = 168 : 12 = 14

d) 100x + \(\frac{100.101}{2}\) = 5750

<=> 100x + 5050 = 5750

=> 100x = 5750 - 5050 = 700

=> x = 700 : 100 = 7

Bài 2:Tìm x

a.6x-15=19               b. 3x+2=20+(-12)

   6x=19+15                  3x+2=8

   6x=34                         3x=6

   x=34:6                         x=6:3

   x=17/3                         x=2

Vậy x=17/3                      Vậy x=2

c) 12x - 33=5.3^3              d) ( x+1 ) + ( x+2 ) + ... + ( x+100 ) = 5750

     12x-33=135                 Vì cứ 1 số hạng lại có 1x

      12x=135-33                 Vậy từ 1 đến 100 có số số hạng là:

      12x=102                           (100-1):1+1=100(số)

       x=102:12                    Tổng dãy số là:

        x=17/2                            (100+1)x100:2=5050

Vậy x=17/2                         Do đó có 100x ta có:

                                           ( x+1 ) + ( x+2 ) + ... + ( x+100 ) = 5750

                                            (x+x+...+x)+(1+2+...+100)=5750

                                             100x+5050=5750

                                              100x=700

                                               x=7

                                               Vậy x=7

a)6x-15=19

6x=19+15

6x=34

x=34/6

b)3x+2=20+(-12)

3x+2=8

3x=8-2

3x=6

x=6/3

x=2

c)12x-33=5*3³=5*27=135

12x=135+33

12x=168

x=168/12

x=14

d)(x+1)+(x+2)+...+(x+100)=5750

(x+x+....+x)+(1+2+3+....+100)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=700/100

x=7

a) 34/6

b) 2

c) 14 

d) 7

Bài 2:Tìm x

a.6x-15=19               b. 3x + 2 =20 + ( - 12 )

   6x=19+15                  3x+2=8

   6x=34                        3x=6

   x=34:6                        x=6:3

   x=17/3                        x=2

Vậy x=17/3                     Vậy x=2

c. 12x - 33=5.3^3              d. ( x+1 ) + ( x+2 ) + ... + ( x+100 ) = 5750

     12x-33=135                 Vì cứ 1 số hạng lại có 1x

      12x=135-33                 Vậy từ 1 đến 100 có số số hạng là:

      12x=102                           (100-1):1+1=100(số)

       x=102:12                    Tổng dãy số là:

        x=17/2                            (100+1)x100:2=5050

Vậy x=17/2                         Do đó có 100x ta có:

                                           ( x+1 ) + ( x+2 ) + ... + ( x+100 ) = 5750

                                            (x+x+...+x)+(1+2+...+100)=5750

                                             100x+5050=5750

                                              100x=700

                                               x=7

                                               Vậy x=7

\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)

\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)

\(100x-5050=50\)

\(100x=50+5050\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=7\)

\(b,x+\left(1+2+3+...+50\right)=2000\)

 \(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)

\(x+1275=2000\)

\(\Rightarrow x=2000-1275=725\)

c , ( x - 1 ) + ( x - 2 ) + ...... + ( x - 100 ) = 50

( x + x + ........ + ) - ( 1 + 2 + ..... + 100 ) = 50

100x - 5050 = 50

100x = 50 + 5050

100x = 5100

=> x = 5100/100 = 51

\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

<=> (x-1)(3x+1)=0

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

<=> 2(x+3)- x(x+3)=0

<=> (x+3)(2-x)=0

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

<=> x(x-2)+3(x-2)=0

<=> (x-2)(x+3)=0

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

<=> x(x-1)+3(x-1)=0

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

b) \(\text{3x (x-1) + x - 1 = 0}\)

\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

c) \(\text{2(x+3) - x ² - 3x = 0}\)

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Rightarrow\left(2-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d) \(\text{x(x - 2) + 3x - 6 = 0}\)

\(\Rightarrow x(x - 2) + 3(x - 2) = 0\\ \Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

e)

\(\text{4x ² - 4x +1 = 0}\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow x=0,5\)

f) \(\text{x +5x ² = 0}\)

\(\Rightarrow x\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

viết lại câu g đi bạn

`a)4x(x-2)+x-2=0`

`<=>(x-2)(4x+1)=0`

`<=>[(x-2=0),(4x+1=0):}`

`<=>[(x=2),(x=-1/4):}`

Vậy `S={2;-1/4}.`

`b)(3x-1)^3-9=0`

`<=>(3x-1-3)(3x-1+3)=0`

`<=>(3x-4)(3x+2)=0`

`<=>[(3x-4=0),(3x+2=0):}`

`<=>[(x=4/3),(x=-2/3):}`

Vậy `S={4/3;-2/3}.`

`c)x^3-8+(x-2)(x+1)=0`

`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`

`<=>(x-2)(x^2+3x+5)=0`

Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`

`<=>x-2=0`

`<=>x=2`

Vậy `S={2}`

a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)

b)Ta có: \(\left(3x-1\right)^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)

b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)

c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)

a) Ta có: \(4x\left(x-2\right)+x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)

b) Ta có: \(\left(3x-1\right)^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)