$B={\huge{(}}\dfrac{1}{\sqrt{1}}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}{\huge{)}}:\dfrac{\sqrt{x}}{x+\sqrt{x}}\ \ (x>0)\\=\dfrac{x+1}{\sqrt{x}}$ Tìm $x$ để $B=2$
1 câu trả lời
Đáp án:
Để `B=2` thì `\frac{x+1}{sqrtx}=2` ($x>0$)
`<=> x+1=2sqrtx`
`<=> 2sqrtx=x+1`
`<=> 4x=x^2+2x+1`
`<=> 4x-x^2-2x-1=0`
`<=> -x^2+2x-1=0`
`<=> -(x-1)^2=0`
`<=> (x-1)^2=0`
`<=> x-1=0`
`<=> x=1` (tmdkxd)
Vậy `x=1` thì `B=2`