Question

Tìm x

1.x^2-2x-1=0

2.x^2-x-1=0

3.x^2+x-3=0

4.4x^2-4x-1=0

5.4x^2-2x-1=0

6.4x^2-x-1=0

7.2x^2-2x-3=0

8.3x^2+3x-1=0

Answer 1

1)x^2-2x-1=0

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

<=>(x-1)² =2

=>x-1 = \(\pm\sqrt{2}\)

=> x= \(\pm\sqrt{2}\) +1

2) x^2-x-1=0

<=> (x^2-x+1/4) -5/4=0

<=>(x+1/2)²= 5/4

=> x+1/2 = \(\pm\sqrt{\dfrac{5}{4}}\)

=>x=\(\pm\sqrt{\dfrac{5}{4}}\) - 1/2

Answer 2

3)x^2+x-3=0

<=> (x^2 + x + 1/4) -13/4=0

<=>(x+1/2)² = 13/4

=> x+1/2 = \(\sqrt{\dfrac{13}{4}}\)

=> x= \(\sqrt{\dfrac{13}{4}}\) -1/2

4) 4x^2-4x-1=0

<=> (4x^2-4x+1)-2=0

<=>(2x-1)² -2=0

<=> (2x-1)² - \(\left(\sqrt{2}\right)^2\) =0

<=> (2x-1 - \(\sqrt{2}\) ) . (2x-1 +\(\sqrt{2}\) )=0

=> 2x-1-\(\sqrt{2}\) =0 hoặc 2x-1+\(\sqrt{2}\) =0

=> 2x= 1+\(\sqrt{2}\) hoặc 2x= 1 - \(\sqrt{2}\)

=> x=\(\dfrac{1+\sqrt{2}}{2}\) hoặc x=\(\dfrac{1-\sqrt{2}}{2}\)

Answer 3

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