Tìm x biết: a) `(x+3)^3-x(3x+1)^2+2x(2x-1)^2 = 5(x+1)(1-x)` b) `(x+1)^3-(x-1)^3-6(x-1)^2=-10` Rất cần một sự giúp đỡ… 17/07/2021 Bởi Arianna Tìm x biết: a) `(x+3)^3-x(3x+1)^2+2x(2x-1)^2 = 5(x+1)(1-x)` b) `(x+1)^3-(x-1)^3-6(x-1)^2=-10` Rất cần một sự giúp đỡ…
Đáp án: Giải thích các bước giải: $(x+3)^3-x(3x+1)^2+2x(2x-1)^2=5(x+1)(1-x)$ $⇔x^3+9x^2+27x+27-x(9x^2+6x+1)+2x(4x^2-4x+1)=-5(x-1)(x+1)$ $⇔x^3+9x^2+27x+27-9x^2-6x^2-x+8x^3-8x^2+2x=-5(x^2-1)$ $⇔(x^3+8x^3-9x^3)+(9x^2-6x^2-8x^2)+(27x-x+2x)+27=-5x^2+5$ $⇔-5x^2+5x^2+28x=5-27$ $⇔28x=-22$ $⇔x=\dfrac{-11}{14}$ $(x+1)^3-(x-1)^3-6(x-1)^2=-10$ $⇔x^3+3x^2+3x+1-x^3+3x^2-3x+1-6(x^2-2x+1)=-10$ $⇔(x^3-x^3)+(3x^2+3x^2)+(3x-3x)+(1x+1)-6x^2+12x-6=-10$ $⇔6x^2+2-6x^2+12x-6=-10$ $⇔(6x^2-6x^2)+12x=-10+6-2$ $⇔12x=-6$ $⇔x=\dfrac{-1}{2}$ Bình luận
Đáp án: a/ $x=-\dfrac{11}{14}$ b/ $x=-\dfrac{1}{2}$ Giải thích các bước giải: a/ $(x+3)^3-x(3x+1)^2+2x(2x-1)^2=5(x+1)(1-x)$ $⇔ x^3+9x^2+27x+27-x(9x^2+6x+1)+2x(4x^2-4x+1)=5(1-x^2)$ $⇔ x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-8x^2+2x-5+5x^2=0$ $⇔ 28x+22=0$ $⇔ x=-\dfrac{22}{28}=-\dfrac{11}{14}$ b/ $(x+1)^3-(x-1)^3-6(x-1)^2=-10$ $⇔ x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+10=0$ $⇔ x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0$ $⇔ 12x+6=0$ $⇔ x=-\dfrac{6}{12}=-\dfrac{1}{2}$ Bình luận
Đáp án:
Giải thích các bước giải:
$(x+3)^3-x(3x+1)^2+2x(2x-1)^2=5(x+1)(1-x)$
$⇔x^3+9x^2+27x+27-x(9x^2+6x+1)+2x(4x^2-4x+1)=-5(x-1)(x+1)$
$⇔x^3+9x^2+27x+27-9x^2-6x^2-x+8x^3-8x^2+2x=-5(x^2-1)$
$⇔(x^3+8x^3-9x^3)+(9x^2-6x^2-8x^2)+(27x-x+2x)+27=-5x^2+5$
$⇔-5x^2+5x^2+28x=5-27$
$⇔28x=-22$
$⇔x=\dfrac{-11}{14}$
$(x+1)^3-(x-1)^3-6(x-1)^2=-10$
$⇔x^3+3x^2+3x+1-x^3+3x^2-3x+1-6(x^2-2x+1)=-10$
$⇔(x^3-x^3)+(3x^2+3x^2)+(3x-3x)+(1x+1)-6x^2+12x-6=-10$
$⇔6x^2+2-6x^2+12x-6=-10$
$⇔(6x^2-6x^2)+12x=-10+6-2$
$⇔12x=-6$
$⇔x=\dfrac{-1}{2}$
Đáp án:
a/ $x=-\dfrac{11}{14}$
b/ $x=-\dfrac{1}{2}$
Giải thích các bước giải:
a/ $(x+3)^3-x(3x+1)^2+2x(2x-1)^2=5(x+1)(1-x)$
$⇔ x^3+9x^2+27x+27-x(9x^2+6x+1)+2x(4x^2-4x+1)=5(1-x^2)$
$⇔ x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-8x^2+2x-5+5x^2=0$
$⇔ 28x+22=0$
$⇔ x=-\dfrac{22}{28}=-\dfrac{11}{14}$
b/ $(x+1)^3-(x-1)^3-6(x-1)^2=-10$
$⇔ x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+10=0$
$⇔ x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0$
$⇔ 12x+6=0$
$⇔ x=-\dfrac{6}{12}=-\dfrac{1}{2}$