Q=\((\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}^3-\sqrt{y}^3}{y-x}):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
1) rút gọn Q
2) chứng minh Q\(\ge0\)
3) so sánh Q với \(\sqrt{Q}\)
Bài 1: Xét \(Q=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a) Rút gọn \(Q\)
b) Chứng minh rằng \(Q\ge0\)
c) So sánh \(Q\) và \(\sqrt{Q}\)
\(Q=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}^3-\sqrt{y}^3}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}\)
1) rút gọn M
2) chứng minh Q\(\ge\)0
3) so sánh Q với \(\sqrt{Q}\)
Giúp mình với <3
B = \(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a, Rút gọn B
b, Chứng minh : B > hoặc = 0
c, So sánh B với \(\sqrt{B}\)
\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right)\)rút gọn biết x=2-\(\sqrt{3}\)và y =\(2+\sqrt{3}\)
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)
Rút gọn:
a/ \(\frac{\left(\sqrt{x^2+9}-3\right)\left(\sqrt{x^2+9}+3\right)\left(x+\sqrt{xy}+y\right)\sqrt{x-2\sqrt{xy}+y}}{x\left(x\sqrt{x}-y\sqrt{y}\right)}\) (với x>0, y\(\ge\)0, x\(\ne\)y
b/ \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)(với x>0 và x\(\ne\)1
c/ \(\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)(với x>0 và x\(\ne\)1
Rút gọn biểu thức: A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right)\)
Với \(x=2-\sqrt{3}\)
\(y=2+\sqrt{3}\)
Cho biểu thức
A= \(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3-\sqrt{y^3}}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a, Rút gọn A
Chứng minh A>0
cho biểu thức: \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\) \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+1}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right).\backslash\ \)với \(x,y\ge0;x,y\ne1\)
a) Rút gọn P
b) Tính P khi \(x=\sqrt[3]{4-2\sqrt{6}}+\sqrt[3]{4+2\sqrt{6}}\)và \(y=x^2+6\)
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)