\(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\left(\sqrt{x}-\sqrt{y^{ }}\right)^2\)
Biết \(0< x\le y\)và \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(x+2y\right)}\right)+\left(\frac{y}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)=\frac{5}{3}\)
Tính \(\frac{x}{y}\)
\(\frac{\left(\sqrt{X}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)
rút gọn:
a)\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\times\left(1+\frac{1}{x}\right)\)
b)\(\left(\frac{2\sqrt{xy}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+\sqrt{y}}\right)\times\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
c)\(\left(\frac{x-1}{\sqrt{x}-1}+\frac{x\sqrt{x}-1}{1-x}\right)\div\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}+1}\)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
C/m các biể thức sau ko phụ thuộc và giá trị của biến thuộc ĐKXĐ:
a) \(\left(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
b) \(\left(\frac{\sqrt{x}}{\sqrt{xy-y}}+\frac{2\sqrt{x}+\sqrt{y}}{\sqrt{xy}-x}\right).\frac{x\sqrt{y}-y\sqrt{x}}{x+2\sqrt{xy}+y}\)
Giải giùm mình bài này
$\left[\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{y-x}\right].\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}$
\(\)
\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right):\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\left(\sqrt{x}+\sqrt{y}\right)-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right].\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
Rút gọn: \(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y}\right).\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{x\sqrt{x}+y\sqrt{y}}\)
Lời giải:
ĐK: \(x,y\geq 0; x\neq y\). Để cho gọn đặt \(\sqrt{x}=a; \sqrt{y}=b\). Khi đó:
\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y}\right).\frac{(\sqrt{x}-\sqrt{y})^2}{x\sqrt{x}+y\sqrt{y}}\)
\(=(\frac{a^2-b^2}{a-b}-\frac{a^3-b^3}{a^2-b^2}).\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{(a^2-b^2)(a+b)-(a^3-b^3)}{a^2-b^2}.\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{ab(a-b)}{(a-b)(a+b)}.\frac{(a-b)^2}{a^3+b^3}=\frac{ab(a-b)^2}{(a+b)(a^3+b^3)}\)
\(=\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})(x\sqrt{x}+y\sqrt{y})}\)
Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)
các bn giải tiếp cho mk bài này vs
\(D=\left\{\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right\}.\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
*** Lưu ý: { ... } là dấu ngoặc vuông nha tại máy mk ko viết dc ngoặc vuông nên viết tạm thành ngoặc nhọn
D = \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}\) . \(\frac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) = \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(\frac{\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{y}\left(\sqrt{y}+2\right)-2\sqrt{xy}+1}{\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)+\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)
\(=\frac{x-2\sqrt{x}+y+2\sqrt{y}-2\sqrt{xy}+1}{x-2\sqrt{xy}+y-1}\)\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1}{\left(\sqrt{x}-\sqrt{y}\right)^2-1}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}-1\right)^2}{\left(\sqrt{x}-\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)}=\frac{\sqrt{x}-\sqrt{y}-1}{\sqrt{x}-\sqrt{y}+1}\)