Question

câu 9,
B=\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{8\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}-x-3}{x-1}-\frac{1}{\sqrt{x}-1}\right)\)
a, rút gọn B
b,tính giá trị của B khi x=3+\(2\sqrt{2}\)
c,chúng minh B≤1với mọi giá trị của m thỏa mãn x≥0,x≠1

Answer 1

\(x\ge0;x\ne1\)

\(B=\left(\frac{\left(\sqrt{x}+1\right)^2}{x-1}-\frac{\left(\sqrt{x}-1\right)^2}{x-1}-\frac{8\sqrt{x}}{x-1}\right):\left(\frac{-x+\sqrt{x}-3}{x-1}-\frac{\sqrt{x}+1}{x-1}\right)\)

\(B=\left(\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)-8\sqrt{x}}{x-1}\right):\left(\frac{-x-4}{x-1}\right)\)

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

Khi \(x=3+2\sqrt{2}\Rightarrow B=\frac{4\sqrt{3+2\sqrt{2}}}{7+2\sqrt{2}}=\frac{4\left(\sqrt{2}+1\right)}{7+2\sqrt{2}}\)

c/ Ta có: \(B-1=\frac{4\sqrt{x}}{x+4}-1=\frac{4\sqrt{x}-x-4}{x+4}=\frac{-\left(\sqrt{x}-2\right)}{x+4}\)

Do \(x\ge0\Rightarrow x+4>0\Rightarrow\frac{-\left(\sqrt{x}-2\right)}{x+4}\le0\)

\(\Rightarrow B-1\le0\Rightarrow B\le1\)