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b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)
Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)
\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
Kl: x=19, y= 5, z=1890
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)
=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)
=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)
Dấu = xảy ra khi x=y=z=6căn 2
`[\sqrt{27}-\sqrt{15}]/[3-\sqrt{5}]+4/[2+\sqrt{3}]-6/\sqrt{3}`
`=[\sqrt{3}(3-\sqrt{5})]/[3-\sqrt{5}]+[4(2-\sqrt{3})]/[4-3]-[2\sqrt{3}.\sqrt{3}]/\sqrt{3}`
`=\sqrt{3}+8-4\sqrt{3}-2\sqrt{3}`
`=8-5\sqrt{3}`
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`[x-y]/[\sqrt{x}+\sqrt{y}]-[x\sqrt{y}+y\sqrt{x}]/\sqrt{xy}` `ĐK: x,y > 0`
`=[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]/[\sqrt{x}+\sqrt{y}]-[\sqrt{xy}(\sqrt{x}+\sqrt{y})]/\sqrt{xy}`
`=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}`
`=-2\sqrt{y}`
Ta có: +) \(3=\left(\sqrt[3]{2}\right)^3+1^3=\left(\sqrt[3]{2}+1\right)\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)\Rightarrow\frac{1}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\frac{\sqrt[3]{2}+1}{3}\)\(\Rightarrow\frac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\sqrt[3]{2}+1\)hay \(x=\sqrt[3]{2}+1\)
+) \(3=\left(\sqrt[3]{4}\right)^3-1^3=\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)\)\(\Rightarrow\sqrt[3]{16}+\sqrt[3]{4}+1=\frac{3}{\sqrt[3]{4}-1}\Rightarrow4+\sqrt[3]{4}+\sqrt[3]{16}=\frac{3\sqrt[3]{4}}{\sqrt[3]{4}-1}\)\(\Rightarrow\frac{6}{4+\sqrt[3]{4}+\sqrt[3]{16}}=\frac{6\sqrt[3]{4}-6}{3\sqrt[3]{4}}=2-\frac{2}{\sqrt[3]{4}}=2-\sqrt[3]{2}\)hay \(y=2-\sqrt[3]{2}\)
Từ đó suy ra \(x+y=\sqrt[3]{2}+1+2-\sqrt[3]{2}=3\)là một số tự nhiên (đpcm)
Ta có: \(x=\frac{3\left(1+\sqrt[2]{2}\right)}{\left(\sqrt[3]{2^2}-\sqrt[3]{2}+1\right)\left(1+\sqrt[3]{2}\right)}=\frac{3\left(1+\sqrt[2]{2}\right)}{1+\left(\sqrt[3]{2}\right)^3}=1+\sqrt[2]{2}\)
\(y=\frac{6\left(2-\sqrt[3]{2}\right)}{\left(2^2+2\sqrt[3]{2}+\sqrt[3]{2^2}\right)\left(2-\sqrt[3]{2}\right)}=\frac{6\left(2-\sqrt[3]{2}\right)}{2^3-\left(\sqrt[3]{2}\right)^3}=2-\sqrt[3]{2}\)
Vậy x+y=1+\(\sqrt[3]{2}+2-\sqrt[3]{2}=3\)là 1 số tự
nhiên
\(x=\dfrac{3}{\sqrt[3]{4}-\sqrt[3]{2}+1}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{\left(\sqrt[3]{2}+1\right)\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\dfrac{3\left(\sqrt[3]{2}+1\right)}{3}=3\sqrt[3]{2}+1\)
\(y=\dfrac{6}{\sqrt[3]{16}+\sqrt[3]{4}+4}=\dfrac{6}{\sqrt[3]{4}\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)}=\dfrac{6\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}.3}=\dfrac{2\left(\sqrt[3]{4}-1\right)}{\sqrt[3]{4}}=\dfrac{2\sqrt[3]{4}}{\sqrt[3]{4}}-\dfrac{\sqrt[3]{8}}{\sqrt[3]{4}}=2-\sqrt[3]{2}\)
=> x + y = \(\sqrt[3]{2}+1+2-\sqrt[3]{2}=3\)