Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

hay \(x\in\left\{-5;2\right\}\)

Rối mắt , loạn thần kinh toàn là x không

Nhiều quá bạn ơi 

a) 2x(x - 5) - x(3 + 2x) = 26

    2x² - 10x - 3x - 2x² = 26

    -10x - 3x = 26

    -13x = 26 => x = -2

b) 5x(x - 1) = x - 1

    5x(x - 1) - (x - 1) = 0

    (x - 1)(5x - 1) = 0

    x - 1 = 0 => x = 1

    5x - 1 = 0 => x = \(\frac{1}{5}\)

Vậy x = 0; x = \(\frac{1}{5}\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

b) \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{1}{5}\end{array}\right.\)

c) \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)

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

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

\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-\frac{2}{3}\end{array}\right.\)

e) \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)

f) \(x^3+x^2-4x=4\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)

\(1,x^3-3x^2=0\)

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

\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)

\(2,3x^3-48x=0\)

\(3x\left(x^2-16\right)=0\)

\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)

\(3,5x\left(x-1\right)=x-1\)

\(5x^2-5x=x-1\)

\(5x^2-6x+1=0\)

\(5x^2-5x-x+1=0\)

\(5x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(5x-1\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

\(4,2\left(x+5\right)-x^2-5x=0\)

\(2x+10-x^2-5x=0\)

\(-x^2-3x+10=0\)

\(-x^2-5x+2x+10=0\)

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

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

\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)

\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x-26=0\)

\(-13\left(x+2\right)=0\)

\(x=-2\left(TM\right)\)

Trả lời:

1, \(x^3-3x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Vậy x = 0; x = 3 là nghiệm của pt.

2, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.

3, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = 1; x = 1/5 là nghiệm của pt.

4, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)

Vậy x = - 5; x = 2 là nghiệm của pt.

5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = - 2 là nghiệm của pt.

a. \(2x\left(x+5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2+10x-3x-2x^2=26\Leftrightarrow7x=26\Leftrightarrow x=\dfrac{26}{7}\)

Vậy \(x=\dfrac{26}{7}\)

b. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

c. \(2\left(x+5\right)-x^2-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

d. \(\left(2x-3\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

e. \(3x^3-48x=0\Leftrightarrow3x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)

f. \(x^3+x^2-4x=4\Leftrightarrow x^3+x^2-4x-4=0\Leftrightarrow\left(x^2-4x+4\right)+\left(x^3-8\right)=0\Leftrightarrow\left(x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)=0\Leftrightarrow\left(x-2\right)\left(x-2+x^2+2x+4\right)=0\left(x-2\right)\left(x^2+3x+2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2x+2\right)=0\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=-2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

g. \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

h. \(x^2-4x+8=2x-1\Leftrightarrow x^2-4x+8-2x+1=0\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy \(x=3\)

__________________________Chúc bạn học tốt____________________________

a) \(\Rightarrow x\left(x+3\right)=0\)

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

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

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

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

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

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

\(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)

2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = - 26 : 13

x = - 2

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) ( x + 2 ) ( x + 3 ) - ( x - 2 ) ( x + 5 ) = 0

     x² + 3x + 2x + 6 - x² + 5x - 2x - 10 = 0

     8x - 4                                              = 0 

    8x                                                    = 0 + 4

     8x                                                  = 4

       x                                                  = 4 : 8 

       x                                                 = 1/2

b) 2x ( x -5 ) - x ( 3 + 2x ) = 26

 2x² - 10x - 3x - 2x²          = 26

-13x                                  = 26

x                                       = 26 : (-13)

x                                      = -2