a) \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

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

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

\(\Leftrightarrow2x-1=-3\Leftrightarrow x=-1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{6}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=-\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)

b: Ta có: \(\left(x-2\right)^2=1\)

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

c: Ta có: \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

a. \(\left(x-\dfrac{1}{2}\right)^2=0\)

<=> \(x-\dfrac{1}{2}=0\)

<=> \(x=\dfrac{1}{2}\)

b. (x - 2)² = 1

<=> (x - 2)² - 1² = 0

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

<=> (x - 3)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)

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

c. (2x - 1)³ = -8

<=> (2x - 1)³ + 2³ = 0

<=> (2x - 1 + 2)\(\left[\left(2x-1\right)^2-2\left(2x-1\right)+2^2\right]=0\)

<=> (2x + 1)(4x² - 4x + 1 - 4x + 2 + 4) = 0

<=> (2x + 1)(4x² - 8x + 7) = 0

<=> \(\left[{}\begin{matrix}2x+1=0\\4x^2-8x+7=0\end{matrix}\right.\)

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

d. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

<=> \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{\left(\sqrt{6}\right)^2}=0\)

<=> \(\left(x+\dfrac{1}{2}-\dfrac{1}{\sqrt{6}}\right)\left(x+\dfrac{1}{2}+\dfrac{1}{\sqrt{6}}\right)=0\)

<=> \(\left(x+\dfrac{3-\sqrt{6}}{6}\right)\left(x+\dfrac{3+\sqrt{6}}{6}\right)=0\)

<=> \(\left[{}\begin{matrix}x+\dfrac{3-\sqrt{6}}{6}=0\\x+\dfrac{3+\sqrt{6}}{6}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=\dfrac{-3-\sqrt{6}}{6}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow x^2-2x+1+9-x^2=0\)

\(\Rightarrow2x=10\Rightarrow x=5\)

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

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

\(\Rightarrow-\left(x+3\right)\left(3x-1\right)=0\)

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

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow x^2-2x+1+9-x^2=0\\ \Leftrightarrow-2x=-10\\ \Leftrightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow x^2-4x+4-4x^2-4x-1=0\\ \Leftrightarrow-3x^2-8x+3=0\\ \Leftrightarrow3x^2+8x-3=0\\ \Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Bài 3: 

b: \(x^2+2x+1=\left(x+1\right)^2\)

c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)

d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)

\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)

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

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

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

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

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

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

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

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

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

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

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`

a,x(x-2)+x-2=0

⇔ (x-2)(x+1)=0

⇔ x=2;x=-1

b,x³+x²+x+1=0

⇔ x²(x+1)+x+1=0

⇔ (x+1)(x²+1)=0

⇔ x=-1

\(x^2-2x=0\)

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

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

b.\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12+\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x^2-4\right)\)

\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)

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

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm

\(a,x^2-2x=0\)

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

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

\(b,\dfrac{x+1}{x-2}-\dfrac{5}{x-2}=\dfrac{12}{x^2-4}+1\) (ĐKXĐ : x ≠ 2 ; x ≠ -2)

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2+3x+2-5x-10=12+x^2+2x-2x+4\)

\(\Leftrightarrow2x=24\)

\(\Leftrightarrow x=12\left(N\right)\)

câu c chưa học :vv

a)

<=> x (x-2 ) = 0

<=> x =0 

x = 2

b)

đkxđ : x khác 2 , x khác -2

<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{x^2-4}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\dfrac{x^2+3x+2}{....}-\dfrac{5x-10}{....}-\dfrac{12}{...}+\dfrac{x^2-4}{....}=0\)

<=> \(x^2+3x+2-5x+10-12+x^2-4=0\)

<=> \(2x^2-2x-4=0\)

<=> x =2 (ktm)

Vậy..

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

= x(x - 1 - 2y)(x - 1 + 2y)

Bài 3:

a: \(2x\left(x-3\right)-x+3=0\)

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

=>(x-3)(2x-1)=0

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

b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

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

=>-x-2=0

=>-x=2

=>x=-2

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

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

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

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

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

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

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

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

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