e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0

\(\Leftrightarrow\) (x - 3) ( x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

    \(\Leftrightarrow\) x = 3                            \(\Leftrightarrow\) x = 2

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

c) Ta có: \(x^3-5x^2-x+5=0\)

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

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

d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)

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

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

a) \(\left(x-4\right)^2-\left(x-4\right)=0\)

\(\left(x-4\right)\left(x-4-1\right)=0\)

\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)

\(\left(x-7\right)\left(5x^2+7\right)=0\)

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

\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)

\(x=7\)

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

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

\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)

a) (x - 4)^2=(x - 4)

(x - 4) (x -4)=(x -4 )

(x - 4) (x - 4)-(x - 4)=0

(x-4) (x-4-1)=0

(x-4) (x-5)=0

TH1:x-4=0                          TH2:x-5=0

            x=4                                      x=5

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

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

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

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

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

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

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)