\(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
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) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
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)\(\dfrac{\sqrt{x+\sqrt{y}}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x-\sqrt{y}}}{2\left(\sqrt{x+\sqrt{y}}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x-\sqrt{y}}}\)
chứng minh
\(\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}=4\)
\(P=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\):\(\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
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 x,y,z>0 và x+y+z=\(\sqrt{2}\). chứng minh rằng
\(A=\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\dfrac{\sqrt{y+z}}{x}+\dfrac{\sqrt{z+x}}{y}+\dfrac{\sqrt{x+y}}{z}\right)\ge4\sqrt{2}\)
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.
\(\left(\sqrt{X}+\dfrac{Y-\sqrt{XY}}{\sqrt{X}+\sqrt{Y}}\right):\left(\dfrac{X}{\sqrt{XY}+y}+\dfrac{Y}{\sqrt{XY}-x}-\dfrac{X+y}{\sqrt{XY}}\right)\)
Cho biểu thức \(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\) Tìm các giá trị x, y, nguyên để P có giá trị bằng 2
(\(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)):\(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
\(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x^3}-\sqrt{y^3}}{x+\sqrt{xy}+y}-2\sqrt{y}\)
\(\left(1-\dfrac{4\sqrt{x}}{x-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\) ĐKXĐ: x>0 ; x≠1 ; x≠4
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\) ĐKXĐ: x>0 và x≠4