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\(\left(a\sqrt{b+1}+b\sqrt{a+1}\right)^2\le\left(a^2+b^2\right)\left(a+b+2\right)=a+b+2\le\sqrt{2\left(a^2+b^2\right)}+2=2+\sqrt{2}\)
\(\Rightarrow a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{2+\sqrt{2}}\)
1.
\(a^2=3-2\sqrt{2}=\sqrt{9}-\sqrt{9-1}\)
2.
\(A=\left(x+y+1-2\sqrt{xy}-2\sqrt{x}+2\sqrt{y}\right)+\left(x-4\sqrt{x}+4\right)+2015\)
\(A=\left(\sqrt{x}-\sqrt{y}-1\right)^2+\left(\sqrt{x}-2\right)^2+2015\ge2015\)
\(A_{min}=2015\) khi \(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
1) \(1019x^2+18y^4+1007z^2\)
\(=\left(15x^2+15y^4\right)+\left(3y^4+3z^2\right)+\left(1004x^2+1004z^2\right)\)
\(\ge2\sqrt{15x^2.15y^4}+2\sqrt{3y^4.3z^2}+2\sqrt{1004x^2.1004z^2}=30xy^2+6y^2z+2008xz\left(đpcm\right)\)
\(a\sqrt{b}-b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(7\sqrt{7}+3\sqrt{3}=\left(\sqrt{7}+\sqrt{3}\right)\left(7-\sqrt{21}+3\right)=\left(\sqrt{7}+\sqrt{3}\right)\left(10-\sqrt{21}\right)\)
\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)
\(1-a\sqrt{a}=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)\)
\(x^2-\sqrt{x}=\sqrt{x}\left(x\sqrt{x}-1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(\left(\sqrt{2}+1\right)^2-4\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\left(\sqrt{5}+2\right)^2-8\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
2 cái trên đều áp dụng HĐT \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
\(5\sqrt{2}-2\sqrt{5}=\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)\)
\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=9\)
\(B=\frac{2}{2}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)
\(B>2\left(\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+...+\frac{\sqrt{36}-\sqrt{35}}{\left(\sqrt{36}-\sqrt{35}\right)\left(\sqrt{36}+\sqrt{35}\right)}\right)\)
\(B>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)
\(B>2\left(\sqrt{36}-1\right)=10>9=A\)
\(\Rightarrow B>A\)
Để biểu thức B có nghĩa thì \(xy\ne0\)
Khi đó ta có:
\(x^3+y^3=2x^2y^2\)
\(\Leftrightarrow\left(x^3+y^3\right)^2=4x^4y^4\)
\(\Leftrightarrow x^6+y^6+2x^3y^3=4x^4y^4\)
\(\Leftrightarrow x^6+y^6-2x^3y^3=4x^4y^4-4x^3y^3\)
\(\Leftrightarrow\left(x^3-y^3\right)^2=4x^4y^4\left(1-\frac{1}{xy}\right)\)
\(\Leftrightarrow1-\frac{1}{xy}=\left(\frac{x^3-y^3}{2x^2y^2}\right)^2\)
\(\Rightarrow\sqrt{1-\frac{1}{xy}}=\left|\frac{x^3-y^3}{2x^2y^2}\right|\) là một số hữu tỉ