a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a, (\(x-2\))² - (2\(x\) + 3)² = 0

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

     (-\(x\) - 5)(3\(x\) +1) = 0

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

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

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

Vậy \(x\in\) { -5;- \(\dfrac{1}{3}\)}

b, 9.(2\(x\) + 1)² - 4.(\(x\) + 1)² = 0 

    {3.(2\(x\) + 1) - 2.(\(x\) +1)}{ 3.(2\(x\) +1) + 2.(\(x\) +1)} = 0

    (6\(x\) + 3 - 2\(x\) - 2)(6\(x\) + 3 + 2\(x\) + 2) = 0

      (4\(x\) + 1)(8\(x\) + 5) =0

        \(\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{8}\end{matrix}\right.\)

          S = { - \(\dfrac{5}{8}\); \(\dfrac{-1}{4}\)}

d, \(x^2\)(\(x\) + 1) - \(x\) (\(x+1\)) + \(x\)(\(x\) -1) = 0

      \(x\left(x+1\right)\).(\(x\) - 1) + \(x\)(\(x\) -1) = 0

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

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

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

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

              S = { -2; 0; 1}

e, (\(x\) - 2)²- (\(x\) - 2)(\(x\) + 2) = 0 

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

      -4 (\(x-2\)) = 0

            \(x\) - 2 = 0

            \(x\) = 2

          S ={ 2}

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

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

b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)

\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)

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

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

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

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

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

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

Do \(\left(x+1\right)^2+1>0\)

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

a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x+3=0\)

hay x=-3

a)4x²-9=0

⇔ (2x-3)(2x+3)=0

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

b)(x+5)²-(x-1)²=0

⇔ (x+5-x+1)(x+5+x-1)=0

⇔ 12(x+2)=0

⇔ x=-2

c)x²-6x-7=0

⇔ x²-7x+x-7=0

⇔ x(x-7)+(x-7)=0

⇔ (x-7)(x+1)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

d)(x+1)²-(2x-1)²=0

⇔ (x+1-2x+1)(x+1+2x-1)=0

⇔3x(2-x)=0

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

a, 4x² - 9 = 0

<=> 4x² = 9

<=> x² = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)

b, (x + 5 )² - ( x - 1 )² = 0

<=> ( x+5-x+1 )(x+5+x-1) = 0

<=> 6(2x+4) = 0

<=> 12x+24=0

<=> 12x = -24

<=> x = -2

c, x²-6x-7=0

<=> x²+x-7x-7=0

<=> x(x+1)-7(x+1)=0

<=> (x-7)(x+1)=0

=> x+7=0 hoặc x+1=0

+ x-7=0 => x=7

+ x+1=0 => x=-1

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

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

<=> (-x+2).3x=0

=> x=0 hoặc (-x+2).3=0

+ (-x+2).3=0 => -3x+6=0 => x=-2

b) (x +5)² -(x -1)²=0

<=> [(x +5) -(x -1)][(x +5) +(x -1)]=0

<=> (x +5 -x +1)(x +5 +x -1)=0

<=> 6(2x+4)=0 <=>12(x +2)=0

=> x +2=0=> x=-2

vậy x= -2

c) x² -6x -7=0

<=> x² -7x +x -7=0

<=> (x² +x)( -7x -7)=0

<=> x(x +1).-7(x +1)=0

<=> (x +1)(x -7)=0

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

Vậy S={-1; 7}

d) (x +1)² -(2x -1)²=0

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

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

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

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

Vậy S={0; \(\dfrac{2}{3}\)}