Bài 2:
\(x-\sqrt{xy}-2y=0\\ \Leftrightarrow x+\sqrt{xy}-2\sqrt{xy}-2y=0\\ \Leftrightarrow\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-2\sqrt{y}\right)=0\\ \Leftrightarrow\sqrt{x}-2\sqrt{y}=0\left(\sqrt{x}+\sqrt{y}>0\right)\\ \Leftrightarrow x=4y\\ \Leftrightarrow Q=\dfrac{\left(4y\right)^3+y^3}{\left(4y+2y\right)^3}=\dfrac{65y^3}{216y^3}=\dfrac{65}{216}\)
Bài 3:
\(\dfrac{x}{x^2+x+1}=2019\Leftrightarrow\dfrac{x^2+x+1}{x}=\dfrac{1}{2019}\\ \Leftrightarrow x+\dfrac{1}{x}+1=\dfrac{1}{2019}\Leftrightarrow x+\dfrac{1}{x}=-\dfrac{2018}{2019}\\ \dfrac{1}{Q}=\dfrac{x^4-x^2+1}{x^2}=x^2+\dfrac{1}{x^2}-1=\left(x+\dfrac{1}{x}\right)^2-3\\ \Leftrightarrow\dfrac{1}{Q}=\dfrac{2018^2}{2019^2}-3=\dfrac{2018^2-3.2019^2}{2019^2}\\ \Leftrightarrow Q=\dfrac{2019^2}{2018^2-3.2019^2}\)
Bài 4:
\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\cdot\dfrac{x+y+z}{xyz}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
Không áp dụng vào bài toán đc, ta phải cm cái khác:
Ta có \(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{a}+\dfrac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\left(1+\dfrac{1}{a}\right)^2-2\cdot\dfrac{a+1}{a}\cdot\dfrac{1}{a+1}+\dfrac{1}{\left(a+1\right)^2}}\\ =\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=\left|1+\dfrac{1}{a}-\dfrac{1}{a+1}\right|\)
Áp dụng vào bài toán:
\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =2008+\dfrac{1}{2}-\dfrac{1}{2010}=...\)
