Question

[Trần Tích Thường]

Tìm x : 

1) \(x^2-4=0\)                                                                        7) \(x^2-14x+13=0\)
2) \(2x^2-8=0\)                                                                   8) \(x^2+10x+16=0\)
3) \(\left(x+3\right)^2=4\)
4) \(\left(x-7\right)^2=36\)

5) \(x^2-14x=-49\)
6) \(x^2+6x+5=0\)

Answer 1

[Huỳnh Quang Sang]

1) x² - 4 = 0

=> x² = 4

=> x = \(\pm\)2

2) 2x² - 8 = 0

=> 2x² = 8

=> x² = 4

=> x = \(\pm2\)

3) (x + 3)² = 4 => (x + 3)² = 2²

=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

4) (x - 7)² = 36

=> (x - 7)² = 6²

=> \(\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

5) x² - 14x = -49

=> x² - 14x + 49 = 0

=> x² - 7x - 7x + 49 = 0

=> x(x - 7) - 7(x - 7) = 0

=> (x - 7)² = 0

=> x = 7

6) x² + 6x + 5 = 0

=> x² + x + 5x + 5 = 0

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

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

=> \(\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

7) x² - 14x + 13 = 0

=> x² - x - 13x + 13 = 0

=> x(x - 1) - 13(x - 1) = 0

=> (x - 1)(x - 13) = 0

=> \(\orbr{\begin{cases}x=1\\x=13\end{cases}}\)

8) x² + 10x +16 = 0

=> x² + 2x + 8x + 16 = 0

=> x(x + 2) + 8(x + 2) = 0

=> (x + 2)(x + 8) = 0

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)