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

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_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_

tớ ra kết quả là 2+\(\frac{5\sqrt{xy}}{x-\sqrt{xy}+y}\) mà thấy số xấu quá :(

ĐKXĐ:

\(P=\left(\frac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{x\sqrt{x}-y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\frac{\left(\sqrt{x}+\sqrt{y}\right)}{x-\sqrt{xy}+y}\)

\(=\left(\frac{x\sqrt{y}-y\sqrt{x}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\right)\)

\(=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}.\frac{1}{\left(x-\sqrt{xy}+y\right)}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

\(P=\frac{\sqrt{xy}}{x-\sqrt{xy}+\frac{y}{4}+\frac{3y}{4}}=\frac{\sqrt{xy}}{\left(\sqrt{x}-\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}}\)

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow P\ge0\) \(\forall x;y\)