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a/ \(Q=\sqrt{x}+\sqrt{y}\)
b/ \(\hept{\begin{cases}x+y=2015\\xy=2016\end{cases}}\)
\(Q^2=x+y+2\sqrt{xy}=2015+2\sqrt{2016}\)
\(\Rightarrow Q=\sqrt{2015+2\sqrt{2016}}\)
Q \(=\left(\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}-xy\right):\left(x\sqrt{x}-y\sqrt{x}-x\sqrt{y}+y\sqrt{y}\right)\)
Q\(=\left(x^2-xy+y^2-xy\right):\left[\sqrt{x}\left(x-y\right)-\sqrt{y}\left(x-y\right)\right]\)
Q\(=\left(x^2-2xy+y^2\right):\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\)
Q \(=\left(x-y\right)^2:\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\)
Q \(=\left(x-y\right):\left(\sqrt{x}-\sqrt{y}\right)\)
Q \(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right):\left(\sqrt{x}-\sqrt{y}\right)\)
Q \(=\sqrt{x}+\sqrt{y}\)
chịu thua vô điều kiện xin lỗi nha : v
muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v
a.\(DK:x,y>0\)
Ta co:
\(A=\frac{x+y+2\sqrt{xy}}{xy}.\frac{\sqrt{xy}\left(x+y\right)}{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)
b.
Ta lai co:
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\ge\frac{2\sqrt{\sqrt{x}.\sqrt{y}}}{4}=1\)
Dau '=' xay ra khi \(x=y=4\)
Vay \(A_{min}=1\)khi \(x=y=4\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
a/
\(P=\left(\frac{x^3+y^3}{x+y}-xy\right):\left(x\sqrt{x}-y\sqrt{x}-x\sqrt{y}+y\sqrt{y}\right)\)
\(=\left(\frac{x^3+y^3-xy\left(x+y\right)}{x+y}\right):\left(\sqrt{x}\left(x-y\right)-\sqrt{y}\left(x-y\right)\right)\)
\(=\left(x-y\right)^2.\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}=\sqrt{x}+\sqrt{y}\)
b/ Áp dụng vi-et ta có: \(\left\{\begin{matrix}x+y=2015\\xy=2016\end{matrix}\right.\)
\(P=\sqrt{x}+\sqrt{y}\)
\(\Rightarrow P^2=x+y+2\sqrt{xy}\)
\(=2015+2\sqrt{2016}\)
\(\Rightarrow P=\sqrt{2015+2\sqrt{2016}}\)