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

Nên \(\sqrt{x^2+3}-x=\sqrt{y^2+3}+y\)

\(\Leftrightarrow\sqrt{x^2+3}-\sqrt{y^2+3}=x+y\) (1)

tương tự

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

nên \(\sqrt{y^2+3}-y=\sqrt{x^2+3}+x\)

\(\Leftrightarrow\sqrt{y^2+3}-\sqrt{x^2+3}=x+y\) (2)

từ (1) và (2) => \(x+y=0\)

\(\left(\left(9\cdot9+9\right):9\right)^2=100\)

99 + (9 : 9)
=> 99 + 1 = 100

Câu hỏi của marivan2016 - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo nhé!

\(2\sqrt{x}+\sqrt{y}=1\Rightarrow\left(2\sqrt{x}+\sqrt{y}\right)^2=1\Leftrightarrow4x+4\sqrt{xy}+y=1\)

Mặt khác \(\left(\sqrt{x}-2\sqrt{y}\right)^2\ge0\forall xy\Leftrightarrow x-4\sqrt{xy}+4y\ge0\)

=>\(\left(4x+4\sqrt{xy}+y\right)+\left(x-\sqrt{4xy}+4y\right)\ge1+0\)

=>\(5\left(x+y\right)\ge1\)

=>\(x+y\ge\frac{1}{5}\)

Dấu "=" xảy ra khi x=4/25 ; y=1/25

x² - 2x + 2010 > 0

=> y >= 0

Ta tìm min(\(\frac{1}{y}\))

\(\frac{1}{y}\)= \(\frac{x^2-2x+2010}{x^2}\)

= \(\frac{\frac{x^2}{2010}-2x+2010+\frac{2009x^2}{2010}}{x^2}\)

= \(\frac{\frac{x^2}{2010}-2x+2010}{x^2}\)+ \(\frac{2009}{2010}\)

= \(\frac{\left(\frac{x}{\sqrt{2010}}-\sqrt{2010}\right)^2}{x^2}\)+ \(\frac{2009}{2010}\)>= \(\frac{2009}{2010}\)

=> Min(_{\(\frac{1}{y}\)}) = \(\frac{2009}{2010}\)khi x = 2010

=> Max(y) = \(\frac{2010}{2009}\) khi x = 2010