Giải phương trình:(2x^2-3x+1)(2x^2+5x+1)=9x^2 07/11/2021 Bởi Liliana Giải phương trình:(2x^2-3x+1)(2x^2+5x+1)=9x^2
Đáp án: `S={\frac{2±\sqrt{2}}{2};\frac{-3±\sqrt{7}}{2}}` Giải thích các bước giải: $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ $⇔[(2x^2+x+1)-4x][(2x^2+x+1)+4x]=9x^2$ $⇔(2x^2+x+1)^2-16x^2=9x^2$ $⇔(2x^2+x+1)^2-25x^2=0$ $⇔(2x^2+x+1-5x)(2x^2+x+1+5x)=0$ $⇔(2x^2-4x+1)(2x^2+6x+1)=0$ $⇔\left[ \begin{array}{l}2x^2-4x+1=0(1)\\2x^2+6x+1=0(2)\end{array} \right.$ `(1)⇔x^2-2x+\frac{1}{2}=0⇔x^2-2x+1=\frac{1}{2}` `⇔(x-1)^2=\frac{1}{2}⇔x-1=\frac{±\sqrt{2}}{2}⇔x=\frac{2±\sqrt{2}}{2}` `(2)⇔x^2+3x+\frac{1}{2}=0⇔x^2+3x+\frac{9}{4}=\frac{7}{4}` `⇔(x+\frac{3}{2})^2=\frac{7}{4}⇔x+\frac{3}{2}=\frac{±\sqrt{7}}{2}⇔x=\frac{-3±\sqrt{7}}{2}` Bình luận
Cách giải: $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ $\to (2x^2+x+1-4x)(2x^2+x+1+4x)=9x^2$ $\to (2x^2+x+1)^2-16x^2=9x^2$ $\to (2x^2+x+1)^2=25x^2$ $\to (2x^2+x+1+5x)(2x^2+x+1-5x)=0$ $\to (2x^2+6x+1)(2x^2-4x+1)=0$ $\to (x^2+3x+\dfrac{1}{2})(x^2-2x+\dfrac{1}{2})=0$ $\to (x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{7}{4})(x^2-2.x.1-\dfrac{1}{2})=0$ $\to [(x+\dfrac{3}{2})^2-\dfrac{7}{4}][(x-1)^2-\dfrac{1}{2}]=0$ $\to (x+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2})(x+\dfrac{3}{2}+\dfrac{\sqrt{7}}{2})(x-1-\dfrac{\sqrt{2}}{2})(x-1+\dfrac{\sqrt{2}}{2})=0$ $\to \left[ \begin{array}{l}x=\dfrac{-3-\sqrt{7}}{2}\\x=\dfrac{-3+\sqrt{7}}{2}\\x=\dfrac{2+\sqrt{2}}{2}\\x=\dfrac{2-\sqrt{2}}{2}\end{array} \right.$ Bình luận
Đáp án: `S={\frac{2±\sqrt{2}}{2};\frac{-3±\sqrt{7}}{2}}`
Giải thích các bước giải:
$(2x^2-3x+1)(2x^2+5x+1)=9x^2$
$⇔[(2x^2+x+1)-4x][(2x^2+x+1)+4x]=9x^2$
$⇔(2x^2+x+1)^2-16x^2=9x^2$
$⇔(2x^2+x+1)^2-25x^2=0$
$⇔(2x^2+x+1-5x)(2x^2+x+1+5x)=0$
$⇔(2x^2-4x+1)(2x^2+6x+1)=0$
$⇔\left[ \begin{array}{l}2x^2-4x+1=0(1)\\2x^2+6x+1=0(2)\end{array} \right.$
`(1)⇔x^2-2x+\frac{1}{2}=0⇔x^2-2x+1=\frac{1}{2}`
`⇔(x-1)^2=\frac{1}{2}⇔x-1=\frac{±\sqrt{2}}{2}⇔x=\frac{2±\sqrt{2}}{2}`
`(2)⇔x^2+3x+\frac{1}{2}=0⇔x^2+3x+\frac{9}{4}=\frac{7}{4}`
`⇔(x+\frac{3}{2})^2=\frac{7}{4}⇔x+\frac{3}{2}=\frac{±\sqrt{7}}{2}⇔x=\frac{-3±\sqrt{7}}{2}`
Cách giải:
$(2x^2-3x+1)(2x^2+5x+1)=9x^2$
$\to (2x^2+x+1-4x)(2x^2+x+1+4x)=9x^2$
$\to (2x^2+x+1)^2-16x^2=9x^2$
$\to (2x^2+x+1)^2=25x^2$
$\to (2x^2+x+1+5x)(2x^2+x+1-5x)=0$
$\to (2x^2+6x+1)(2x^2-4x+1)=0$
$\to (x^2+3x+\dfrac{1}{2})(x^2-2x+\dfrac{1}{2})=0$
$\to (x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{7}{4})(x^2-2.x.1-\dfrac{1}{2})=0$
$\to [(x+\dfrac{3}{2})^2-\dfrac{7}{4}][(x-1)^2-\dfrac{1}{2}]=0$
$\to (x+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2})(x+\dfrac{3}{2}+\dfrac{\sqrt{7}}{2})(x-1-\dfrac{\sqrt{2}}{2})(x-1+\dfrac{\sqrt{2}}{2})=0$
$\to \left[ \begin{array}{l}x=\dfrac{-3-\sqrt{7}}{2}\\x=\dfrac{-3+\sqrt{7}}{2}\\x=\dfrac{2+\sqrt{2}}{2}\\x=\dfrac{2-\sqrt{2}}{2}\end{array} \right.$