4. Cho a;b khác 0.C/m rằng

T

transformers123

Đặt $\dfrac{a}{b}+\dfrac{b}{a}=x \rightarrow x^2=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+2$ (ĐK: $x \ge 2$)
bất đẳng thức trên dược viết lại:
$x^2+2 \ge 3x$
$(x-2)(x-1) \ge 0$
vì $x \ge 2$ nên: $x-2 \ge 0;\ x-1 > 0$
$\Longrightarrow (x-2)(x-1) \ge 0$ (luôn đúng)
Ok=))
 
M

manhnguyen0164

Đặt $\dfrac{a}{b}+\dfrac{b}{a}=x \rightarrow x^2=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+2$ (ĐK: $x \ge 2$)
bất đẳng thức trên dược viết lại:
$x^2+2 \ge 3x$
$(x-2)(x-1) \ge 0$
vì $x \ge 2$ nên: $x-2 \ge 0;\ x-1 > 0$
$\Longrightarrow (x-2)(x-1) \ge 0$ (luôn đúng)
Ok=))

Sai rồi bạn ơi... chỗ $x \ge 2$ lỡ a,b trái dấu thì sao..a,b phải cùng dương dấu ms dc...đầu bài phải xét a,b trái dấu hoặc cùng dấu
 
S

songdzianhem

Èm hèm

\Leftrightarrow[TEX](\frac{x}{y}+\frac{y}{x}-1)(\frac{x}{y}+\frac{y}{x}-2)\Leftrightarrow\frac{[(x-\frac{y}{2})^2+\frac{3y^2}{4}](x-y)^2}{x^2y^2}[/TEX]

đpcm
 
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V

vitaminngoc

[tex]\frac{7}{8} qwertyuioplkjhgfdsazxcvbnm tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt[/tex]
 
M

manhnguyen0164

[tex]\frac{7}{8} qwertyuioplkjhgfdsazxcvbnm tttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt[/QUOTE] Bà ơi không gõ được $LaTeX$ rồi lại vào đây phá...không đăng bài thì đừng phá nhá bà.....spam khó chịu quá[/tex]
 
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