Đáp án :
Nếu `(\sqrt{1+x}+\sqrt{1-x})/(\sqrt{1+x}-\sqrt{1-x})=\sqrt{2}` thì `x=(±2.\sqrt{2})/3`
Giải thích các bước giải :
`(\sqrt{1+x}+\sqrt{1-x})/(\sqrt{1+x}-\sqrt{1-x})=\sqrt{2}`
`<=>((\sqrt{1+x}+\sqrt{1-x})/(\sqrt{1+x}-\sqrt{1-x}))^2=(\sqrt{2})^2`
`<=>(\sqrt{1+x}+\sqrt{1-x})^2/(\sqrt{1+x}-\sqrt{1-x})^2=2`
`<=>(1+x+2.\sqrt{1+x}.\sqrt{1-x}+1-x)/(1+x-2.\sqrt{1+x}.\sqrt{1-x}+1-x)=2`
`<=>(2+2.\sqrt{1+x}.\sqrt{1-x})/(2-2.\sqrt{1+x}.\sqrt{1-x})=2`
`<=>(2+2.\sqrt{1+x}.\sqrt{1-x})/(2-2.\sqrt{1+x}.\sqrt{1-x})-2=0`
`<=>(2+2.\sqrt{1+x}.\sqrt{1-x})/(2-2.\sqrt{1+x}.\sqrt{1-x})-(2(2-2.\sqrt{1+x}.\sqrt{1-x}))/(2-2.\sqrt{1+x}.\sqrt{1-x})=0`
`<=>(2+2.\sqrt{1+x}.\sqrt{1-x}-4+4.\sqrt{1+x}.\sqrt{1-x})/(2-2.\sqrt{1+x}.\sqrt{1-x})=0`
`<=>(6.\sqrt{1+x}.\sqrt{1-x}-2)/(2-2.\sqrt{1+x}.\sqrt{1-x})=0`
`<=>6.\sqrt{1+x}.\sqrt{1-x}-2=0`
`<=>6.\sqrt{1+x}.\sqrt{1-x}=2`
`<=>\sqrt{1+x}.\sqrt{1-x}=1/3`
`<=>(\sqrt{1+x}.\sqrt{1-x})^2=(1/3)^2`
`<=>(1+x).(1-x)=1/9`
`<=>1-x^2=1/9`
`<=>x^2=1-1/9`
`<=>x^2=9/9-1/9`
`<=>x^2=8/9`
`<=>x^2=(4.2)/9`
`<=>x^2=((±2.\sqrt{2})/3)^2`
`<=>x=(±2.\sqrt{2})/3`
Vậy : Nếu `(\sqrt{1+x}+\sqrt{1-x})/(\sqrt{1+x}-\sqrt{1-x})=\sqrt{2}` thì `x=(±2.\sqrt{2})/3`
