ĐK : \(2-x^2\ge0\Leftrightarrow x^2\le2\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\)
Có \(\dfrac{1}{x}+\dfrac{1}{\sqrt{2-x^2}}\ge\dfrac{4}{x+\sqrt{2-x^2}}=\dfrac{4}{1.x+1.\sqrt{2-x^2}}\)
\(\ge\dfrac{4}{\sqrt{\left(1^2+1^2\right)\left(x^2+2-x^2\right)}}=\dfrac{4}{2}=2\)
Dấu "=" xảy ra <=> \(\dfrac{1}{x}=\dfrac{1}{\sqrt{2-x^2}}\Leftrightarrow x=\sqrt{2-x^2}\)
<=> \(\left\{{}\begin{matrix}x^2=2-x^2\\-\sqrt{2}\le x\le\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\-\sqrt{2}\le x\le\sqrt{2}\end{matrix}\right.\Leftrightarrow x=\pm1\)
Vậy \(x=\pm1\)nghiệm phương trình
Mk sửa lại bài trước
ĐK : \(2-x^2>0\Leftrightarrow x^2< 2\Leftrightarrow-\sqrt{2}< x< \sqrt{2}\)
Với x > 0 Có \(\dfrac{1}{x}+\dfrac{1}{\sqrt{2-x^2}}\ge\dfrac{4}{x+\sqrt{2-x^2}}=\dfrac{4}{1.x+1.\sqrt{2-x^2}}\)
\(\ge\dfrac{4}{\sqrt{\left(1^2+1^2\right)\left(x^2+2-x^2\right)}}=\dfrac{4}{2}=2\)
Dấu "=" xảy ra <=> \(\dfrac{1}{x}=\dfrac{1}{\sqrt{2-x^2}}\Leftrightarrow x=\sqrt{2-x^2}\)
<=> \(\left\{{}\begin{matrix}x^2=2-x^2\\-\sqrt{2}< x< \sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\-\sqrt{2}< x< \sqrt{2}\end{matrix}\right.\Leftrightarrow x=1\)
Với x < 0 => \(\dfrac{1}{x}+\dfrac{1}{\sqrt{2-x^2}}< \dfrac{1}{\sqrt{2}}< 2\left(\text{loại}\right)\)
Vậy x = 1nghiệm phương trình
\(\Leftrightarrow\dfrac{1}{x}-1+\dfrac{1}{\sqrt{2-x^2}}-1=0\)
\(\Leftrightarrow\dfrac{1-x}{x}+\dfrac{1-\sqrt{2-x^2}}{\sqrt{2-x^2}}=0\Rightarrow\left(1-x\right)\sqrt{2-x^2}+x-x\sqrt{2-x^2}=0\)
\(\Rightarrow\sqrt{2-x^2}-x\sqrt{2-x^2}+x-x\sqrt{2-x^2}=0\)
\(\Leftrightarrow\sqrt{2-x^2}-2x\sqrt{2-x^2}+x=0\)
\(\Leftrightarrow\sqrt{2-x^2}-1-\left(2x\sqrt{2-x^2}-2\right)+x-1=0\)
\(\Leftrightarrow\dfrac{2-x^2-1}{\sqrt{2-x^2}+1}-\dfrac{4x^2\left(2-x^2\right)-4}{2x\sqrt{2-x^2}+2}+x-1=0\)
\(\Leftrightarrow\dfrac{1-x^2}{\sqrt{2-x^2}+1}+\dfrac{8x^2-4x^4-4}{2x\sqrt{2-x^2}+2}+x-1=0\)
\(\Leftrightarrow\dfrac{\left(1-x\right)\left(x+1\right)}{\sqrt{2-x^2}+1}+\dfrac{-4\left(x^4-2x^2+1\right)}{2x\sqrt{2-x^2}+2}+x-1=0\)
\(\Leftrightarrow\dfrac{\left(1-x\right)\left(x+1\right)}{\sqrt{2-x^2}+1}-\dfrac{4\left(x^2-1\right)^2}{2x\sqrt{2-x^2}+2}+x-1=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
