Bài 1: Bạn tự đặt điều kiện nhé
[TEX]PT \Leftrightarrow \frac{1}{2008x+1}+ \frac{1}{2011x+5}= \frac{1}{2009x+2}+ \frac{1}{2010x+4}[/TEX]
[TEX]\Leftrightarrow \frac{4019x+6}{(2008x+1)(2011x+5)}= \frac{4019x+6}{(2009x+2)(2010x+4)}[/TEX]
Đến đây giải OK rồi
Bài 2:
Để ý rằng::
[TEX]( \sqrt{2010}y+ \sqrt{1+2010y^2})( \sqrt{1+2010y^2}- \sqrt{2010}y)=1[/TEX]
[TEX] \sqrt{2010}y+ \sqrt{1+2010y^2} > \sqrt{2010}y+ \sqrt{2010}|y| \geq 0[/TEX]
[TEX]\Rightarrow gt \Leftrightarrow \sqrt{2009}x+ \sqrt{1+2009x^2}= \sqrt{1+2010y^2}- \sqrt{2010}y[/TEX]
[TEX]\Leftrightarrow \sqrt{2009}x+\sqrt{2010}y+ \sqrt{1+2009x^2}- \sqrt{1+2010y^2}=0[/TEX]
[TEX]\Leftrightarrow \sqrt{2009}x+\sqrt{2010}y+ \frac{2009x^2-2010y^2}{ \sqrt{1+2009x^2}+ \sqrt{1+2010y^2}}=0[/TEX]
[TEX]\Leftrightarrow \sqrt{2009}x+\sqrt{2010}y=0(1),hoac:1+ \frac{ \sqrt{2009}x- \sqrt{2010}y}{ \sqrt{1+2009x^2}+ \sqrt{1+2010y^2}}=0(2)[/TEX]
Ta có:
[TEX] \sqrt{1+2009x^2}+ \sqrt{1+2010y^2} > \sqrt{2009}|x|+ \sqrt{2010}|y| \geq | \sqrt{2009}x- \sqrt{2010}y|[/TEX]
[TEX]\Rightarrow \frac{| \sqrt{2009}x- \sqrt{2010}y|}{ \sqrt{1+2009x^2}+ \sqrt{1+2010y^2}} \leq 1[/TEX]
[TEX]\Rightarrow 1+ \frac{ \sqrt{2009}x- \sqrt{2010}y}{ \sqrt{1+2009x^2}+ \sqrt{1+2010y^2}} > 1-1=0[/TEX]
Từ (1) \Rightarrow ĐPCM
đề nói CM x/y mà sao bạn lại làm thế ??
bài 2 ák,,,,,,,,,,,,,,,,,,,,,,,,,,,