Question

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

tim x, tui can gap nhe

Answer 1

\(1+\left(2x-1\right)\left(4x-5\right)-4x^2=0\)

\(1+8x^2-14x+5-4x^2=0\)

\(4x^2-14x+6=0\)

\(2x^2-7x+3=0\)

\(2x^2-6x-x+3=0\)

\(2x\left(x-3\right)-\left(x-3\right)=0\)

\(\left(2x-1\right)\left(x-3\right)=0\)

\(2x-1=0\) hoặc \(x-3=0\)

\(2x=1\) hoặc \(x-3=0\)

\(x=\frac12\) hoặc \(x=3\)

Vậy \(x=\frac12\) hoặc \(x=3\)

Answer 2

1 + (2x - 1)(4x - 5) - 4x² = 0

1 + 8x² - 10x - 4x + 5 - 4x² = 0

4x² - 14x + 6 = 0

4x² - 2x - 12x + 6 = 0

(4x² - 2x) - (12x - 6) = 0

2x(2x - 1) - 6(2x - 1) = 0

(2x - 1)(2x - 6) = 0

2x - 1 = 0 hoặc 2x - 6 = 0

*) 2x - 1 = 0

2x = 1

*) 2x - 6 = 0

2x = 6

x = 6 : 2

x = 3

Vậy:

  

Answer 3

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

⇔ 4x^2 - 14x + 6 = 0

⇔ 2x^2 - 7x + 3 = 0

⇔ (2x - 1)(x - 3) = 0

⇔ x = 1/2 hoặc x = 3

Vậy x ∈ {1/2; 3}