cho Q=\(\dfrac{1+\sqrt{x}}{\sqrt{x}}\) 

a) so sánh Q với 1

A=\(\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x-4\sqrt{x}+4}\right)\)\(\dfrac{x\sqrt{x}-2x-4\sqrt{x}-8}{\sqrt{x}}\)

A=\(\left(\dfrac{1}{3-2\sqrt{2}}-\dfrac{6}{\sqrt{2}+2}\right)\)\(\left(3+5\sqrt{2}\right)\)

a)A=\(\dfrac{1}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a>\(\dfrac{1}{2}\)     

b)A=\(\dfrac{\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}-1}\)+\(\dfrac{\sqrt{x+2\sqrt{x-1}}}{\sqrt{x-1+1}}\) với x>2 

c)\(\dfrac{a+b}{b^2}\)\(\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}\) với a+b>0; b≠0

d)A=\(\left(\sqrt{\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1

e)A=\(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)}{\left(x-1\right)^4}}\) với x≠1; y≠1; y>o

f)A=\(\sqrt{\dfrac{m}{1-2x+x^2}}\)\(\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) với m>0; x≠4

g)A=\(\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\)\(\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\) với x>0; x≠4

h)\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)\(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1

giải phương trình sau:

a)\(\sqrt{x^2-9}\) - 3\(\sqrt{x-3}\) =0       b)\(\sqrt{4x^2-12x+9}\) =x - 3

c)\(\sqrt{x^2+6x+9}\) =3x-1

a) \(\dfrac{2}{5}\sqrt{25}\) -\(\dfrac{1}{2}\sqrt{4}\)       b)0,5\(\sqrt{0,09}\) +5\(\sqrt{0,81}\)       c)\(\dfrac{2}{5}\sqrt{\dfrac{25}{36}}\) -\(\dfrac{5}{2}\sqrt{\dfrac{4}{25}}\)

d)-2\(\sqrt{\dfrac{-36}{-16}}\) + 5\(\sqrt{\dfrac{-81}{-25}}\)

tìm x để các biểu thức sau có nghĩa:

a)\(\sqrt{\left(x-2\right)}\)+\(\dfrac{1}{x-5}\)    b)\(\sqrt{\left(2x-6\right)\left(7-x\right)}\)    c)\(\sqrt{4x^2-25}\)       

d)\(\dfrac{2}{x^2-9}\)-\(\sqrt{5-2x}\)      e)\(\dfrac{x}{x^2-4}\)+\(\sqrt{x-2}\)

\(\sqrt{\dfrac{-5}{x^2+6}}\) xđk?

\(\dfrac{2-x}{x^2-9}\) xđk?