\(x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=\frac{x^2}{\sqrt{x}}+\frac{y^2}{\sqrt{y}}+\frac{z^2}{\sqrt{z}}\)   (1)

Áp dụng BDT Cauchy-Schwarz: 

\(\left(1\right)\ge\frac{\left(x+y+z\right)^2}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Ta lại có: \(x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}\Leftrightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le3\)

Thay vào ta có \(\left(1\right)\ge\frac{1}{\sqrt{3}}\)

Dấu = xảy ra khi x=y=z=1/3

\(27x^3\sqrt{x}+27y^3\sqrt{y}+27z^3\sqrt{z}+\sqrt{x}+\sqrt{y}+\sqrt{z}\ge6\sqrt{3}\left(x^2+y^2+z^2\right)\)

Lại có: \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\)

Thay vào -> dpcm

\(=\sqrt{2}\sqrt{2-\sqrt{3}}\)\(+\sqrt{2}\sqrt{2+\sqrt{3}}\)

\(=\sqrt{4-2\sqrt{3}}\)\(+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\)\(+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|\)\(+\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+\sqrt{3}+1\)

\(=2\sqrt{3}\)

đặt \(2008=a\)

\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\sqrt{\left(a+1\right)^2-\frac{2\left(a+1\right).a}{a+1}+\left(\frac{a}{a+1}\right)^2}=\)\(\sqrt{\left(a+1-\frac{a}{a+1}\right)^2}=a+1-\frac{a}{a+1}\)=2008+1- \(\frac{2008}{2009}\)

=> A= 2008+1 = 2009

Cái đề bài kiểu j vậy ? Đùa nhau à!

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= 15 + 12 - 30

= -3

\(\left(\sqrt{75}+\sqrt{16}-\sqrt{300}\right)\sqrt{3}\)

\(=3\left(\sqrt{25}+\sqrt{16}-\sqrt{100}\right)\)

\(=3\left(5+4-10\right)\)

\(=-3\)

\(\left(x+3\right).\sqrt{26-x^2}=\left(x-6\right).\left(x+3\right) \)
\(\Leftrightarrow\sqrt{26-x^2}=x-6\)
\(\Leftrightarrow26-x^2=\left(x-6\right)^2\)
\(\Leftrightarrow26-x^2=x^2-12x+36\)\(\Leftrightarrow2x^2-12x+10=0\)
\(\Leftrightarrow\left(x-5\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
 vậy x= 5 hoặc x=1

Ta có:

\(n^2\equiv0;1\left(mod4\right)\)

\(\Rightarrow2^{n^2}\equiv1;2\left(mod5\right);2^{4n^4+1-n^2}\equiv2;1\left(mod5\right)\)

\(\Rightarrow2^{n^2}+2^{4n^4+1-n^2}\equiv3\left(mod5\right)\)

\(\Rightarrow2^{n^2}+2^{4n^4+1-n^2}=5k+3\left(k\in N\right)\)

\(\Rightarrow2^{M\left(n\right)}-8=2^{5k+3}-8=2^{5k}.2^3-8\equiv8-8\equiv0\left(mod31\right)\)