Question

Bài 6: Giải các phương trình sau:

1)

2)

3)

4)

5)

6)

7)

8) 

9)

10)

11)

12)

13)

14) x² – 2x + 1 = 0

15) 1 + 3x + 3x² + x³ = 0

Answer 1

Bài 6: 

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

\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)

\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)

\(\Leftrightarrow-14x-9=0\)

\(\Leftrightarrow-14x=9\)

\(\Leftrightarrow x=-\dfrac{9}{14}\)

Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)

Answer 2

`1)2x(x-5)-(x+3)^2=3x-x(5-x)`

`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`

`<=>x^2-16x-9=x^2-2x`

`<=>14x=-9`

`<=>x=-9/14`