ĐKXĐ x>1
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x-1}\right)\left(\sqrt{x+1}\right)}{\sqrt{x}-1}\\ < =>A=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\left(\sqrt{x}+1\right)\\ < =>A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ < =>A=x-\sqrt{x}+1\\ x=5+4\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{4-3}}\\ < =>x=5+4\left(2-\sqrt{3}\right)\\ < =>x=13-4\sqrt{3}=12-2\cdot2\sqrt{3}+1\\ < =>x=\left(2\sqrt{3}-1\right)^2\\ \)thay x vào A mới rút gọn, ta được:
A=\(13-4\sqrt{3}-\sqrt{\left(2\sqrt{3}-1\right)^2}+1\\ < =>A=13-4\sqrt{3}-2\sqrt{3}+1+1\\ < =>A=15-6\sqrt{3}\)
x=\(5+4\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)=\(5+\sqrt{\dfrac{4-2\sqrt{3}}{2+\sqrt{3}}}=5+\left(\sqrt{3}-1\right)\sqrt{\dfrac{1}{2+\sqrt{3}}}\)
=\(5+\left(\sqrt{3}-1\right)\sqrt{2-\sqrt{3}}\)=\(5+\sqrt{6-3\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
=\(5+\dfrac{\sqrt{12-6\sqrt{3}}-\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
=\(5+\dfrac{3-\sqrt{3}-\sqrt{3}+1}{\sqrt{2}}\)=\(5+\dfrac{4-2\sqrt{3}}{\sqrt{2}}=\dfrac{5\sqrt{2}+4-2\sqrt{3}}{\sqrt{2}}=5+2\sqrt{2}-\sqrt{6}\)
lại có : A=\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
=\(\dfrac{x\sqrt{x}-1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
=\(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
=\(\sqrt{x}-1-2\sqrt{x}+1+2\sqrt{x}+2=\sqrt{x}\)
thay x vào A (tự tính)
