Question

a, Tính A =\(\dfrac{1}{\sqrt{3}+2}-\sqrt{7+4\sqrt{3}}\)

b, Rút gọn ; B=\(\dfrac{x+2\sqrt{x}+1}{x-1}-\dfrac{2x}{x-\sqrt{x}}\) [ x lớn hơn 0 , x\(\ne1\) ]

Answer 1

a)A=\(\dfrac{1}{\sqrt{3}+2}-\sqrt{7+4\sqrt{3}}=\dfrac{\sqrt{3}-2}{\left(\sqrt{3}\right)^2-2^2}-\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}=2-\sqrt{3}-\sqrt{\left(2+\sqrt{3}\right)^2}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

b)ĐKXĐ:x>0 x\(\ne1\)

B=\(\dfrac{x+2\sqrt{x}+1}{x-1}-\dfrac{2x}{x-\sqrt{x}}=\dfrac{x\sqrt{x}+2x+\sqrt{x}-2x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}+2x+\sqrt{x}-2x\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-x\sqrt{x}+\sqrt{x}}{\sqrt{x}\left(x-1\right)}=\dfrac{\sqrt{x}\left(1-x\right)}{\sqrt{x}\left(x-1\right)}=-1\)

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