bài 1, giải pt a, 2x^3-x^2-2x+1=0 b,2x^2-x=3-6x c,y^2-6/y = 3x+1 29/08/2021 Bởi Everleigh bài 1, giải pt a, 2x^3-x^2-2x+1=0 b,2x^2-x=3-6x c,y^2-6/y = 3x+1
CHÚC BẠN HỌC TỐT!!! Trả lời: $a,2x^3-x^2-2x+1=0$ $⇔2x^3-2x^2+x^2-x-x+1=0$ $⇔(x-1)(2x^2+x-1)=0$ $⇔(x-1)(2x^2+2x-x-1)=0$ $⇔(x-1)(x+1)(2x-1)=0$ $⇔\left[ \begin{array}{l}x=1\\x=-1\\x=\dfrac{1}{2}\end{array} \right.$ Vậy `S={±1;1/2}`. $b, 2x^2-x=3-6x$ $⇔2x^2+5x-3=0$ $⇔2x^2-x+6x-3=0$ $⇔(2x-1)(x+3)=0$ $⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-3\end{array} \right.$ Vậy `S={1/2;-3}`. $c,$ ĐKXĐ: $x\neq 0$ $\dfrac{x^2-6}{x}=3x+1$ $⇔x^2-6=3x^2+x$ $⇔2x^2+x+6=0$ $⇔x^2+\dfrac{1}{2}x+3=0$ $⇔x^2+2.x.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{47}{3}=0$ $⇔\bigg{(}x+\dfrac{1}{4}\bigg{)}^2=-\dfrac{47}{3}$ (vô lí) $⇒$ Phương trình vô nghiệm. Vậy $S=∅$. Bình luận
\(a) 2x^3-x^2-2x+1=0\\↔x^2(2x-1)-(2x-1)=0\\↔(x^2-1)(2x-1)=0\\↔(x-1)(x+1)(2x-1)=0\\↔x-1=0\quad or\quad x+1=0\quad or\quad 2x-1=0\\↔x=1\quad or\quad x=-1\quad or\quad x=\dfrac{1}{2}\\\text{Vậy pt có tập nghiệm}\,\,S=\{1;-1;\dfrac{1}{2}\}\\b) 2x^2-x=3-6x\\↔2x^2-x+6x-3=0\\↔(2x^2+6x)-(x+3)=0\\↔2x(x+3)-(x+3)=0\\↔(2x-1)(x+3)=0\\↔2x-1=0\quad or\quad x+3=0\\↔x=\dfrac{1}{2}\quad or\quad x=-3\\\text{Vậy pt có tập nghiệm}\,\,S=\{\dfrac{1}{2};-3\}\\c)\dfrac{y^2-6}{y}=3y+1(y\ne 0)\\↔y^2-6=y(3y+1)\\↔y^2-6=3y^2-y\\↔y^2-3y^2-y-6=0\\↔-2y^2-y-6=0\\↔y^2+\dfrac{1}{2}y+3=0\\↔y^2+2.\dfrac{1}{4}.y+\dfrac{1}{16}+\dfrac{47}{16}=0\\↔(y+\dfrac{1}{4})^2+\dfrac{47}{16}=0(vô\,\,lý)\\\text{Vậy pt vô nghiệm}\,\,S=\varnothing\) Bình luận
CHÚC BẠN HỌC TỐT!!!
Trả lời:
$a,2x^3-x^2-2x+1=0$
$⇔2x^3-2x^2+x^2-x-x+1=0$
$⇔(x-1)(2x^2+x-1)=0$
$⇔(x-1)(2x^2+2x-x-1)=0$
$⇔(x-1)(x+1)(2x-1)=0$
$⇔\left[ \begin{array}{l}x=1\\x=-1\\x=\dfrac{1}{2}\end{array} \right.$
Vậy `S={±1;1/2}`.
$b, 2x^2-x=3-6x$
$⇔2x^2+5x-3=0$
$⇔2x^2-x+6x-3=0$
$⇔(2x-1)(x+3)=0$
$⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-3\end{array} \right.$
Vậy `S={1/2;-3}`.
$c,$ ĐKXĐ: $x\neq 0$
$\dfrac{x^2-6}{x}=3x+1$
$⇔x^2-6=3x^2+x$
$⇔2x^2+x+6=0$
$⇔x^2+\dfrac{1}{2}x+3=0$
$⇔x^2+2.x.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{47}{3}=0$
$⇔\bigg{(}x+\dfrac{1}{4}\bigg{)}^2=-\dfrac{47}{3}$ (vô lí)
$⇒$ Phương trình vô nghiệm.
Vậy $S=∅$.
\(a) 2x^3-x^2-2x+1=0\\↔x^2(2x-1)-(2x-1)=0\\↔(x^2-1)(2x-1)=0\\↔(x-1)(x+1)(2x-1)=0\\↔x-1=0\quad or\quad x+1=0\quad or\quad 2x-1=0\\↔x=1\quad or\quad x=-1\quad or\quad x=\dfrac{1}{2}\\\text{Vậy pt có tập nghiệm}\,\,S=\{1;-1;\dfrac{1}{2}\}\\b) 2x^2-x=3-6x\\↔2x^2-x+6x-3=0\\↔(2x^2+6x)-(x+3)=0\\↔2x(x+3)-(x+3)=0\\↔(2x-1)(x+3)=0\\↔2x-1=0\quad or\quad x+3=0\\↔x=\dfrac{1}{2}\quad or\quad x=-3\\\text{Vậy pt có tập nghiệm}\,\,S=\{\dfrac{1}{2};-3\}\\c)\dfrac{y^2-6}{y}=3y+1(y\ne 0)\\↔y^2-6=y(3y+1)\\↔y^2-6=3y^2-y\\↔y^2-3y^2-y-6=0\\↔-2y^2-y-6=0\\↔y^2+\dfrac{1}{2}y+3=0\\↔y^2+2.\dfrac{1}{4}.y+\dfrac{1}{16}+\dfrac{47}{16}=0\\↔(y+\dfrac{1}{4})^2+\dfrac{47}{16}=0(vô\,\,lý)\\\text{Vậy pt vô nghiệm}\,\,S=\varnothing\)