3x² + 6x + 5 = 0

ta  có 3 (x² + 2x + 1) + 2 = 0

           3( x+ 1)² + 2 = 0

        3(x+1)² ≥ 0 ⇔ 3(x+1)² + 2 ≥ 2 > 0 ∀ x

vậy pt vô nghiệm hay x ϵ \(\varnothing\)

hưng học giỏi

    2x² + 7x + 3 

=   x² + 6x + 9  + x² - 9 + x + 3

= (x + 3)² + (x -3)(x+3) + (x+3)

= (x+3) ( x + 3 + x -3 + 1)

= (x+3)(2x +1)

.

\(x^2-2x+1-49=\left(x-1\right)^2-7^2=\left(x-8\right)\left(x+6\right)\)

x² - 2x - 48

=x² -2x+ 1-49

=(x-1)² - 7²

=(x-1-7). (x-1+7)

=(x-8).(x+6)

    x² - 2x - 8 

= x² - 2x + 1  - 9

= (x-1)² - 9

= (x-1-3)(x-1+3)

= (x-4)(x+2)

b, x² - 2x - 15

=x² - 2x + 1 - 16

= (x-1)² - 4²

= (x-1 -4)(x-1+4)

= (x-5)(x+3)

a, \(=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)

b, \(=x^2-2x+1-16=\left(x-1\right)^2-16=\left(x-5\right)\left(x+3\right)\)

đk x khác 0 ; -1

\(E=\left(\dfrac{x+1+x^2}{x\left(x+1\right)}\right):\dfrac{x}{x^2+x}=\dfrac{x^2+x+1}{x}=x+1+\dfrac{1}{x}\)

\(E=\left(\dfrac{1}{x}+\dfrac{x}{x+1}\right):\dfrac{x}{x^2+x}\left(ĐK:x\ne0;x\ne-1\right)\)

\(E=\left(\dfrac{x+1}{x^2+x}+\dfrac{x^2}{x^2+x}\right):\dfrac{x}{x^2+x}\\ E=\left(\dfrac{x+1}{x^2+x}:\dfrac{x}{x^2+x}\right)+\left(\dfrac{x^2}{x^2+x}:\dfrac{x}{x^2+x}\right)\\ E=\dfrac{2x+x}{x^2+x}\)