rongphuongdong
12-02-2010, 09:45 PM
Bài 1: Cho http://latex.codecogs.com/gif.latex?x^2 + y^2 \leq x+y . Chứng minh: http://latex.codecogs.com/gif.latex?x+y \leq 2
Bài 2: Cho a,b>0. Chứng minh
http://latex.codecogs.com/gif.latex?\frac{a^2}{b^2} + \frac{b^2}{a^2} \geq \frac{a}{b} + \frac{b}{a}
GIẢI:
Bài 1:
http://diendan.hocmai.vn/latex.php?2(x+y)\geq 2(x^2+y^2)\geq (x+y)^2\Rightarrow x+y\leq 2
Bài 2:
http://diendan.hocmai.vn/latex.php?2\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\r ight)\geq \left(\frac ab+\frac ba\right)^2 \geq 2\left(\frac ab+\frac ba\right)
Bài 2: Cho a,b>0. Chứng minh
http://latex.codecogs.com/gif.latex?\frac{a^2}{b^2} + \frac{b^2}{a^2} \geq \frac{a}{b} + \frac{b}{a}
GIẢI:
Bài 1:
http://diendan.hocmai.vn/latex.php?2(x+y)\geq 2(x^2+y^2)\geq (x+y)^2\Rightarrow x+y\leq 2
Bài 2:
http://diendan.hocmai.vn/latex.php?2\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\r ight)\geq \left(\frac ab+\frac ba\right)^2 \geq 2\left(\frac ab+\frac ba\right)