Rút gọn biểu thức:
A = (\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\) - \(\sqrt{xy}\)) + (\(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\))
B = (\(\sqrt{a}\) + \(\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)) : (\(\dfrac{a}{\sqrt{ab}}\) + \(\dfrac{b}{\sqrt{ab-a}}\) - \(\dfrac{a+b}{\sqrt{ab}}\))
C = \(\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\)(\(\dfrac{\sqrt{b}}{a-\sqrt{ab}}\) + \(\dfrac{\sqrt{b}}{a+\sqrt{ab}}\))
D = (\(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\) - \(\dfrac{x\sqrt{x}-y\sqrt{y}}{x-y}\)) . \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{x\sqrt{x}+y\sqrt{y}}\)
\(P=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\):\(\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
a) Với giá trị nào cùa x thì biểu thức có nghĩa
b) Rút gọn P
c) Tím P với x=3 và y=\(\dfrac{2}{2-\sqrt{3}}\)
Giúp với ạ
\(CHO\:A\:,b,c,\:x,y,z,>0\:VA\dfrac{A}{X}=\dfrac{B}{Y}=\dfrac{C}{Z}\:CM:\:\sqrt{AX}+\sqrt{BY}+\sqrt{CZ\:}=\left(\sqrt{A+b+c\:}\right)\:\left(\sqrt{X+y+z}\right)\)
Cho A= \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\)
a) Đặt điều kiện để biểu thức A có nghĩa
b) Rút gọn A
bài 1 :rút gọn
\(\dfrac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
bài 2 ; tính
\(\sqrt{\left(1-2\right)^2}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{1}{2}}\)
Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
Cho biểu thức:
A = (\(\sqrt{x}\) + \(\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)) : (\(\dfrac{x}{\sqrt{xy}+y}\) + \(\dfrac{y}{\sqrt{xy}-x}\) - \(\dfrac{x+y}{\sqrt{xy}}\))
a) Rút gọn A
b) Tính giá trị của biểu thức A biết x = 3; y = 4 + 2\(\sqrt{3}\)
a, \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
b, \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Các bạn giúp mk vs, Mk cần gấp.
a)\(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7-2}}-\dfrac{\sqrt{7-5}}{2}\) =4+\(\sqrt{11-3\sqrt{7}}\)
b)\(\dfrac{\sqrt{x+\sqrt{y}}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x-\sqrt{y}}}{2\left(\sqrt{x+\sqrt{y}}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x-\sqrt{y}}}\)