2x*(x-7)+(3-x)^2 > 3(x+1)^2

Đáp án:

$x<\dfrac{3}{13}$

Giải thích các bước giải:

$2x(x-7)+(3-x)^2>3(x+1)^2$

$⇔2x^2-14x+9-6x+x^2>3(x^2+2x+1)$

$⇔2x^2-14x+9-6x+x^2>3x^2+6x+3$

$⇔26x-6<0$

$⇔x<\dfrac{3}{13}$

`2x(x-7)+(3-x)^2>3(x+1)^2`

`2x^2-14x+9-6x+x^2>3(x^2+2x+1)`

`3x^2-20x+9>3x^2+6x+3`

`3x^2-20x-3x^2-6x>3-9`

`-26x> -6`

`x<3/13`

Vậy `x<3/13`