1, \(x^2\) - 9 = 0

 (\(x\) - 3)(\(x\) + 3) = 0

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

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

 vậy \(x\) \(\in\) {-3; 3}

7, (3\(x\) + 1)² - 16 = 0

    (3\(x\) + 1 - 4)(3\(x\) + 1 + 4) = 0

    (3\(x\) - 3).(3\(x\) + 5) = 0

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

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

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

Vậy \(x\) \(\in\) {1; - \(\dfrac{5}{3}\)}

10, (\(x\) + 3)² - \(x^2\) = 45

      [(\(x\) + 3) - \(x\)].[(\(x\) + 3) + \(x\)] = 45

                 3.(2\(x\) + 3)  = 45

                     2\(x\) + 3   = 15

                     2\(x\)          = 12

                       \(x\)          = 6

`@` `\text {Ans}`

`\downarrow`

`1,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`2,`

`25 - x^2 = 0`

`<=> x^2 = 25 - 0`

`<=> x^2 = 25`

`<=> x^2 = (+-5)^2`

`<=> x = +-5`

Vậy,` S= {5; -5}`

`3,`

`-x^2 + 36 = 0`

`<=> -x^2 = 0 - 36`

`<=> -x^2 = -36`

`<=> x^2 = 36`

`<=> x^2 = (+-6)^2`

`<=> x = +-6`

Vậy, `S= {6; -6}`

`4,`

`4x^2 - 4 = 0`

`<=> 4x^2 = 0+4`

`<=> 4x^2 = 4`

`<=> x^2 = 4 \div 4`

`<=> x^2 = 1`

`<=> x^2 = (+-1)^2`

`<=> x = +-1`

Vậy, `S= {1; -1}`

`@` `\text {Kaizuu lv uuu}`

a: =>25-4x=1

=>4x=24

hay x=6

b: =>2x-4=0

hay x=2

c: =>x-35=115

hay x=150

d: =>x-2=12

hay x=14

e: =>x-36=216

hay x=252

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

a) \(x^2-x=x-1\)

\(\Leftrightarrow x^2-2x+1=0\)

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

\(\Rightarrow x=1\)

b) \(\left(x^2-36\right)-\left(x+6\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy..

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

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

\(\Leftrightarrow-4x+2=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm1\end{matrix}\right.\)

Vậy..

a) x^2 - x = x - 1

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

\(x^2-2.x.1+1^2=0\)

\(\left(x-1\right)^2=0=>x-1=0=>x=1\)
b) (x^2 - 36) - (x+6) = 0

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

=> (x+6)(x-6) -(x+6) =0

=> (x-6)(x+6-1) =0

=> (x-6)(x+5)=0

=> x=6 hoặc x= (-5)
c) (2x-1)^2 - (4x^2 - 1)= 0

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

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

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

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

=> 2x-1=0 => 2x=1 => x= \(\dfrac{1}{2}\)
d) x^2(x^2 - 4) - (x^2 - 4 ) = 0

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

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

=> x=2;-2;1 hoặc (-1)

\(4\left(6-x\right)+x^2-12x+36=0\)

\(24-4x+x^2-12x+36=0\)

\(x^2-16x+60=0\)

\(x^2-2x8+8^2-8^2+60=0\)

\(\left(x-8\right)^2-4=0\)

\(\left(x-8\right)^2=4\)

\(\left(x-8\right)^2=\left(\pm2\right)^2\)

\(\orbr{\begin{cases}x-8=2\Rightarrow x=10\\x-8=-2\Rightarrow x=6\end{cases}}\)

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)

\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)

\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)

d) \(4x^3-36x=0\)

\(\Leftrightarrow4x\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

Vậy...

\(x^3-4x^2-9x+36=0\)

\(x^2\left(x-4\right)-9\left(x-4\right)=0\)

\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)

\(\Rightarrow x-4=0\) hay \(x^2-9=0\)

\(\Rightarrow x=4\) hay \(x^2=9=3^2\)

\(\Rightarrow x=4\) hay \(x=\pm3\)

⇔x²(x-4) -9(x-4) = 0

⇔(x-4).(x-3).(x+3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)