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a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
Bài 1
a) \(A=\left(4-\sqrt{15}\right)\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\sqrt{\left(4-\sqrt{15}\right)\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}.\left(\sqrt{5}+\sqrt{3}\right).\sqrt{2}=\sqrt{\left(4-\sqrt{15}\right).\left(16-15\right).2}.\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{8-2\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{5-2\sqrt{5}.\sqrt{3}+3}.\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}.\left(\sqrt{5}+\sqrt{3}\right)=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)
Ta có công thức tổng quát\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
Vậy \(B=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{16}-\sqrt{15}=\sqrt{16}-\sqrt{1}=4-1=3\)
b) \(6x^4-7x^2-3=0\Leftrightarrow6x^4-9x^2+2x^2-3=0\Leftrightarrow3x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\Leftrightarrow\left(2x^2-3\right)\left(3x^2+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}2x^2-3=0\\3x^2+1=0\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(2x^2-3=0\Leftrightarrow2x^2=3\Leftrightarrow x^2=\frac{3}{2}\Leftrightarrow x=\frac{\pm\sqrt{6}}{2}\)
Vậy S={\(\frac{-\sqrt{6}}{2};\frac{\sqrt{6}}{2}\)}