Question

B=(\(\dfrac{\sqrt{X}+1}{\sqrt{X}-1}\)-\(\dfrac{\sqrt{X}-1}{\sqrt{X}+1}\)-\(\dfrac{8\sqrt{X}}{X-1}\)):(\(\dfrac{\sqrt{X}-X-3}{X-1}\)-\(\dfrac{1}{\sqrt{X}-1}\)) a) rút gọn B b)tính giá trị của B khi x=3+2\(\sqrt{2}\) c)chứng minh rằng B \(\le\)1 VỚI MỌI GIÁ TRỊ CỦA X THỎA MÃN X\(\ge\)0 ; x\(\ne\)1

Answer 1

a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-\left(x+4\right)}=\dfrac{4\sqrt{x}}{x+4}\)

b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)