a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

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

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

a) x = 8 3 .                            b) x = − 9 20 .  

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

(x - 2)(x² + 2x + 4) + 2(x² - 4) - 5(x - 2) = 0

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

(x - 2)[(x+2)² + 2(x+2) - 5]= 0

(x - 2)[(x + 2)² + 2(x + 2) + 1 - 6] = 0

( x - 2)[(x + 2 + 1)² - 6] = 0

(x - 2)[(x + 3)² - 6] = 0

(x - 2)(x + 3 - \(\sqrt{6}\))(x + 3 + \(\sqrt{6}\)) = 0

TH1. x - 2 = 0 <=> x = 2

TH2. x + 3 - \(\sqrt{6}\) = 0 <=> x = \(\sqrt{6}-3\)

TH3. x + 3 + \(\sqrt{6}\) = 0 <=> x = \(-\sqrt{6}-3\)

S = {2; \(\sqrt{6}-3\); \(-\sqrt{6}-3\)}

\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)

b: \(\Leftrightarrow x+3>0\)

hay x>-3

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

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

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+4x+4+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+4+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\\left(x+2\right)^2=2\left(l\right)\end{matrix}\right.\)

Vậy: x=2

a) x² - 2 = 0

x² = 2

x = -√2 (loại) hoặc x = √2 (loại)

Vậy không tìm được x Q thỏa mãn đề bài

b) x² + 7/4 = 23/4

x² = 23/4 - 7/4

x² = 4

x = 2 (nhận) hoặc x = -2 (nhận)

Vậy x = -2; x = 2

c) (x - 1)² = 0

x - 1 = 0

x = 1 (nhận)

Vậy x = 1

bạn linh này giỏi dữ ta

a) x² - 2 = 0

x² = 2

x = √2 hoặc -√2 (loại) (x ϵ Q)

Vậy x ϵ rỗng

b) x² + 7/4 =23/4

x² = 23/4 - 7/4

x² = 16/4 = 4

x² = 4 = (-2)² = 2²

x² = (-2)²

x = -2 (Nhận)

x² = 2

x = 2 (Nhận)

Vậy x ϵ ( 2 , -2 )

c) (x-1)² = 0

x-1 = 0

x = 1

Vậy x = 1