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c: =>3x^2+3y^2=39 và 3x^2-2y^2=-6
=>5y^2=45 và x^2=13-y^2
=>y^2=9 và x^2=4
=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=5\\\sqrt{x}-\sqrt{y}=-\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y}=1+\dfrac{11}{2}=\dfrac{13}{2}\end{matrix}\right.\)
=>x=1 và y=169/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4-3=1\\-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9-2=7\end{matrix}\right.\)
=>x+1=11/9 và y+4=-11/19
=>x=2/9 và y=-87/19
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: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
3.
\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)
\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)
1. ta có : \(4\sqrt{7}=\sqrt{112}\)
\(3\sqrt{3}=\sqrt{27}\)
ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)
2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)
3. \(x^2=\left(3+\sqrt{2}\right)^2\)
\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)
ta thấy : \(x^2=y^2\Rightarrow x=y\)