Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
a, \(P=\left(x^4-8x^3+16x^2\right)+12x^2-48x+35\)
\(=\left(x^2-4x\right)^2+12\left(x^2-4x\right)+36-1\)
\(=\left(x^2-4x+6\right)^2-1\)
\(=\left[\left(x-2\right)^2+2\right]^2-1\)
\(\ge2^2-1=3\)
Cách khác \(P=\left(x-2\right)^2\left[\left(x-2\right)^2+4\right]+3\ge3\)
Đẳng thức xảy ra khi \(x=2.\)
b, \(xy\le\frac{\left(x+y\right)^2}{4}=9\)
Áp dụng bđt Co6si: \(\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{y^2}}=\frac{2}{xy}\)
\(Q\ge\frac{102}{xy}+xy=xy+\frac{81}{xy}+\frac{21}{xy}\ge2\sqrt{xy.\frac{81}{xy}}+\frac{21}{9}=\frac{61}{3}.\)
Dấu bằng xảy ra khi \(x=y=3.\)
\(=\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x^2-y^2\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}\)
\(=\dfrac{\sqrt{xy}\left(x+y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}=-\sqrt{x}+\sqrt{y}\)(1)
Khi x=3 và \(y=4+2\sqrt{3}\) vào (1), ta được:
\(=-\sqrt{3}+\sqrt{4+2\sqrt{3}}=-\sqrt{3}+\sqrt{3}+1=1\)
\(A=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)
\(A=\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)+y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{x\sqrt{xy}-x^2+y\sqrt{xy}+y^2-y^2+x^2}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}=\frac{y+x}{y-x}\)
KO CÓ GIÁ TRỊ y sao tính đây !!!!!!
CÒN \(x=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\) nhé