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Bài 1 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, ĐKXĐ : \(-x^2+10x-25\ge0\)
=> \(x^2-10x+25\le0\)
=> \(\left(x-5\right)^2\le0\)
=> \(x-5\le0\)
=> \(x\le5\)
Bài 2 :
a, Ta có : \(A=\sqrt{\left(2\sqrt{2}-5\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
=> \(A=5-2\sqrt{2}+\sqrt{5}-2=3-2\sqrt{2}+\sqrt{5}\)
b, Ta có : \(B=\sqrt{9+4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
=> \(B=\sqrt{4+2.2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\)
=> \(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
=> \(B=2+\sqrt{5}-\sqrt{5}+1=3\)
c, Ta có : \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
=> \(C=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{1+2\sqrt{3}+3}}{\sqrt{2}}+\frac{\sqrt{1-2\sqrt{3}+3}}{\sqrt{2}}\)
=> \(C=\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{2}}\)
=> \(C=\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
Ta có : \(a^3=10+3\sqrt[3]{\left(5+\sqrt{52}\right)\left(5-\sqrt{52}\right)}\left(\sqrt[3]{5+\sqrt{52}}+\sqrt[3]{5-\sqrt{52}}\right)\)
\(=10+3\sqrt[3]{-27}.a=10-9a\)
\(\Rightarrow a^3+9a-10=0\Rightarrow\left(a-1\right)\left(a^2+a+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a^2+a+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\\left(a+\dfrac{1}{2}\right)^2+\dfrac{39}{4}>0\end{matrix}\right.\)
\(\Rightarrow a=1\) \(\Rightarrow f\left(a\right)=1+1+1^2+.....+1^{2015}=2016\)
\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+3\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\right)\left(\frac{1}{\sqrt[3]{4-\sqrt{15}}}.\sqrt[3]{4-\sqrt{15}}\right)\)
\(+4-\sqrt{15}\)
<=> \(x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3x\)
<=> \(x^3-3x+2006=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=\frac{4+\sqrt{15}}{16-15}+4-\sqrt{15}+2006\)
<=> \(x^3-3x+2006=2014\)