ai giúp em bài này vs ạ $x^{2}$ -3x+1=-$\frac{\sqrt{3} }{3}$ $\sqrt{x^{4}+x^{2}+1}$ 14/07/2021 Bởi Skylar ai giúp em bài này vs ạ $x^{2}$ -3x+1=-$\frac{\sqrt{3} }{3}$ $\sqrt{x^{4}+x^{2}+1}$
Giải thích các bước giải: $x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{x^4+x^2+1}$ $\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{x^4+2x^2+1-x^2}$ $\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{(x^2+1)^2-x^2}$ $\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{(x^2+1-x)(x^2+1+x)}$ $\to 3(x^2-3x+1)=-\sqrt{3}.\sqrt{x^2-x+1}\sqrt{x^2+x+1}$ $\to 3(2(x^2-x+1)-(x^2+x+1))=-\sqrt{3}.\sqrt{x^2-x+1}.\sqrt{x^2+x+1}$ Đặt $\sqrt{x^2+x+1}=a,\sqrt{x^2-x+1}=b$ $\to 3(2a^2-b^2)=-\sqrt{3}ab$ $\to 6a^2+\sqrt{3}ab-3b^2=0$ $\to (3a-b\sqrt{3})(2a+\sqrt{3}b)=0$ $\to 3a-b\sqrt{3}=0$ $\to a\sqrt{3}=b$ $\to 3a^2=b^2$ $\to 3(x^2+x+1)=x^2-x+1$ $\to 2x^2+4x+2=0$ $\to x^2+2x+1=0$ $\to (x+1)^2=0$ $\to x=-1$ Bình luận
Giải thích các bước giải:
$x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{x^4+x^2+1}$
$\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{x^4+2x^2+1-x^2}$
$\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{(x^2+1)^2-x^2}$
$\to x^2-3x+1=-\dfrac{\sqrt{3}}{3}.\sqrt{(x^2+1-x)(x^2+1+x)}$
$\to 3(x^2-3x+1)=-\sqrt{3}.\sqrt{x^2-x+1}\sqrt{x^2+x+1}$
$\to 3(2(x^2-x+1)-(x^2+x+1))=-\sqrt{3}.\sqrt{x^2-x+1}.\sqrt{x^2+x+1}$
Đặt $\sqrt{x^2+x+1}=a,\sqrt{x^2-x+1}=b$
$\to 3(2a^2-b^2)=-\sqrt{3}ab$
$\to 6a^2+\sqrt{3}ab-3b^2=0$
$\to (3a-b\sqrt{3})(2a+\sqrt{3}b)=0$
$\to 3a-b\sqrt{3}=0$
$\to a\sqrt{3}=b$
$\to 3a^2=b^2$
$\to 3(x^2+x+1)=x^2-x+1$
$\to 2x^2+4x+2=0$
$\to x^2+2x+1=0$
$\to (x+1)^2=0$
$\to x=-1$