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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)\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)=\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}\)=\(\sqrt{3}\)
b)
\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)=\(\frac{\sqrt{y}\left(\sqrt{y}-2\right)}{\sqrt{y}-2}\)=\(\sqrt{y}\)
d) \(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}{\sqrt{x}-1}\)=\(\sqrt{x}\)+3
e)\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)=\(\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)=\(\sqrt{y}\)-1
g)\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x+3}}\)
chúc bạn học tốt
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
\(P=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\sqrt{2}\left(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\right)\)
\(=\sqrt{2}\left(\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\right)=\sqrt{2}\)
2/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{y^2-\frac{7}{y^2}}=a\\\sqrt{y-\frac{7}{y^2}}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a^2-b^2=y^2-y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=y\\\left(a-b\right)\left(a+b\right)=y^2-y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=y\\a-b=y-1\end{matrix}\right.\)
\(\Rightarrow b=\frac{1}{2}\Rightarrow\sqrt{y-\frac{7}{y^2}}=\frac{1}{2}\Rightarrow y-\frac{7}{y^2}=\frac{1}{4}\Rightarrow4y^3-y^2-28=0\)
\(\Rightarrow y=2\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)
\(\Rightarrow3x^2-xy-2y^2=0\)
\(\Rightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-\frac{2}{3}y\end{matrix}\right.\) thay vào 1 trong 2 pt là xong