\(x^2=0\)

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

\(\Rightarrow x=0\)

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\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

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

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

\(x^2=0\)

\(\Rightarrow x=0\left(tm\right)\)

Vậy: \(x=0.\)

\(---\)

\(x^2=16\)

\(\Rightarrow x^2=\left(\pm4\right)^2\)

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

Vậy: \(x\in\left\{4;-4\right\}.\)

x²-4²=0

x-4=0

x=0+4

x=4

`x^2-16=0`

`x^2=16`

`@TH1:x^2=4^2`

       `=>x=4`

`@TH2:x^2=(-4)^2`

       `=>x=-4`

Vậy `x=4` hoặc `x=-4`

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

4

-4

-3

\(=x^2+16x+48=x^2+4x+12x+48=\left(x+4\right)\left(x+12\right)\)

=(x+4)(x+12)

c. 2 x =16= 2 4 ⇒ x = 4

d. x 2 =9= 3 2  ⇒ x = 3

− x 2 + 2 x 3 + x + 1 4 + 2 x + 1 6 = 8 3 ⇒ − 6 x + 8 x + 3 x + 1 + 2 2 x + 1 12 = 32 12 ⇒ 9 x + 5 12 = 32 12 ⇒ 9 x + 5 = 32 ⇒ x = 3

a) x = 4

b) x = 3

c) x = 2

d) x = 1

e) x = 3

f) x = 2

g) x = 4

h) x = 3

a)16. x² = 64

x² = 64 : 16

x² = 4

x² = 2²

⇒ x = 2

b) (5.x - 2) - 64 = -36

(5.x - 2) = -36 + 64

5.x - 2 = 28

5.x = 28 + 2

5.x = 30

x = 30 : 5

x = 6

c) (2x - 10).(5 - x) = 0

TH1: 2x - 10 = 0

         2x = 0 + 10

         2x = 10

           x = 10 : 2

           x = 5

TH2: 5 - x = 0

              x = 5 - 0

              x = 5

⇒ Vậy x = 5.

cứu tui

x=0 thử lại có đúng đâu cj

cái đầu là x - 4 á iem, iu tui nhìu quá nên lú hẻ :<<