Question

Giai phuong trinh:

a) \(\sqrt{x^2+6}=x-2.\sqrt{x^2-1}\)

b) \(x^2+3x+1=\left(x+3\right).\sqrt{x^2+1}\)

Answer 1

b/ Đặt \(\sqrt{x^2+1}=a\ge0\)

\(\Rightarrow a^2+3x=\left(x+3\right)a\)

\(\Leftrightarrow\left(3-a\right)\left(x-a\right)=0\)

Answer 2

a/ Dựa vô TXĐ thì thấy \(x< 2\)

\(\Rightarrow\sqrt{x^2+6}+2\sqrt{x^2-1}-x>\sqrt{6}-2>0\)

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Answer 3

alibaba nguyễn

b/ Đặt \sqrt{x^2+1}=a\ge0x2+1​=a≥0

\Rightarrow a^2+3x=\left(x+3\right)a⇒a2+3x=(x+3)a

\Leftrightarrow\left(3-a\right)\left(x-a\right)=0⇔(3−a)(x−a)=0

a/ Dựa vô TXĐ thì thấy x&lt; 2x<2

\Rightarrow\sqrt{x^2+6}+2\sqrt{x^2-1}-x&gt;\sqrt{6}-2&gt;0⇒x2+6​+2x2−1​−x>6​−2>0

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