a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

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

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

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) Ta có: \(x^2-x-12=0\)

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

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

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

Bằng 0 nha bạn

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)

b)\(=>x-1-x-2=0\)

\(=>-3=0\left(vl\right)\) => x ko tồn tại

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

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

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

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

Em hc bảng xét dáu chx ??

a) Ta có: (x-1)(x-4)>0

\(\Leftrightarrow\left[{}\begin{matrix}x-4>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)

b) Ta có: (x-6)(x-7)<0

\(\Leftrightarrow\left\{{}\begin{matrix}x-6>0\\x-7< 0\end{matrix}\right.\Leftrightarrow6< x< 7\)

c) Ta có: \(\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow1\le x\le2\)

d) Ta có: \(\left(x-2\right)\left(x-\dfrac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge0\\x-\dfrac{2}{3}\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le\dfrac{2}{3}\end{matrix}\right.\)

a) Ta có: (x-1)(x-4)>0

⇔[x−4>0x−1<0⇔[x>4x<1⇔[x−4>0x−1<0⇔[x>4x<1

b) Ta có: (x-6)(x-7)<0

⇔{x−6>0x−7<0⇔6<x<7⇔{x−6>0x−7<0⇔6<x<7

c) Ta có: (x−1)(x−2)≤0(x−1)(x−2)≤0

⇔{x−1≥0x−2≤0⇔1≤x≤2⇔{x−1≥0x−2≤0⇔1≤x≤2

d) Ta có: (x−2)(x−23)≥0(x−2)(x−23)≥0

⇔⎡⎣x−2≥0x−23≤0⇔⎡⎣x≥2x≤23

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1