a,B=\(\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{x\sqrt{x}-x}{\sqrt{x}-1}\)(đk:\(x>1\))
=\(\frac{\sqrt{x-1}+\sqrt{x}}{x-1-x}+\frac{\sqrt{x-1}-\sqrt{x}}{x-1-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
=\(-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}+\sqrt{x}+x\)
=\(x-2\sqrt{x-1}\)
b,Có B=\(x-2\sqrt{x-1}=\left(x-1\right)-2\sqrt{x-1}+1=\left(\sqrt{x-1}-1\right)^2\)
=> \(B\ge0\) \(\)vs mọi x >1
Dấu "=" xảy ra<=> \(\sqrt{x-1}=1\)
<=> x-1=1
<=> x=2
=> Để B>0 <=> x>1 và x\(\ne2\)
c ,\(x=\frac{53}{9-2\sqrt{7}}=\frac{53\left(9+2\sqrt{7}\right)}{9^2-\left(2\sqrt{7}\right)^2}=\frac{53\left(9+2\sqrt{7}\right)}{53}=9+2\sqrt{7}\)
Thay x=\(9+2\sqrt{7}\) vào B có:
B=\(9+2\sqrt{7}-2\sqrt{9+2\sqrt{7}-1}\)
=\(9+2\sqrt{7}-2\sqrt{8-2\sqrt{7}}\)
=\(9+2\sqrt{7}-2\sqrt{\left(\sqrt{7}-1\right)^2}\)
=\(9+2\sqrt{7}-2\left|\sqrt{7}-1\right|=9+2\sqrt{7}-2\left(\sqrt{7}-1\right)=9+2\sqrt{7}-2\sqrt{7}+2\)
=11
Vậy B=11