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

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

(x - 5)(x - 1) = 0

x - 5 = 0 hoặc x - 1 = 0

*) x - 5 = 0

x = 5

*) x - 1 = 0

x = 1

Vậy x = 1; x = 5

b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)

x³ - 8 - x³ - 8 = 2x + 4

2x = -8 - 8 - 4

2x = -20

x = -20 : 2

x = -10

a)

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

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

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

\(x-5=0\) hoặc \(x-1=0\)

+) \(x-5=0\\ \Rightarrow x=5\)

+) \(x-1=0\\ \Rightarrow x=1\)

Vậy \(x=1\) hoặc \(x=5\)

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

\(x^3-8-x^3-8=2x+4\)

\(2x=-8-8-4\)

\(2x=-20\)

 \(x=-20:2\)

 \(x=-10\)

Vậy \(x=-10\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

b) \(9x^2-4-\left(3x-2\right)^2=0\)

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

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

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

\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)

\(\Leftrightarrow3x-2=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)

a) \(6x^2-72x=0\)

\(6x\left(x-12\right)=0\)

\(6x=0\) hoặc \(x-72=0\)

*) \(6x=0\)

\(x=0\)

*) \(x-12=0\)

\(x=12\)

Vậy \(x=0;x=12\)

b) \(-2x^4+16x=0\)

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

\(-2x=0\) hoặc \(x^3-8=0\)

*) \(-2x=0\)

\(x=0\)

*) \(x^3-8=0\)

\(x^3=8\)

\(x=2\)

Vậy \(x=0;x=2\)

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

\(x^2-5x-x^2+6x-9=0\)

\(x-9=0\)

\(x=9\)

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

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

\(-6x^2+12x=0\)

\(-6x\left(x-2\right)=0\)

\(-6x=0\) hoặc \(x-2=0\)

*) \(-6x=0\)

\(x=0\)

*) \(x-2=0\)

\(x=2\)

Vậy \(x=0;x=2\)

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)

\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)

\(\Leftrightarrow12x=-2\)

hay \(x=-\dfrac{1}{6}\)

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)

\(\Leftrightarrow2x=-40\)

hay x=-20

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

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

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)

\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-5x-15=0\)

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

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

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

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

c) \(2x^5-4x^3+2x=0\)

\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)

\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)

\(\Rightarrow2x\left(x^2-1\right)^2=0\)

\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)

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

\(\text{#}Toru\)

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}