[Toán 9]biểu thức

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mamy007

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tuyn

Bài 1: ĐK: x \geq 0
[TEX]\Rightarrow x+ \sqrt{x}+1 \geq 1 \Rightarrow (x+ \sqrt{x}+1)^2 \geq x+ \sqrt{x}+1[/TEX]
[TEX]\Rightarrow \frac{4}{(x+ \sqrt{x}+1)^2} \leq \frac{4}{x+ \sqrt{x}+1}[/TEX]
[TEX]\Leftrightarrow P^2 \leq 2P[/TEX]
Dấu "=" xảy ra khi x=0
Bài 2:
Ta có: [TEX]( \sqrt{y^2+3}+y)( \sqrt{y^2+3}-y)=(y^2+3)-y^2=3[/TEX]
[TEX] \sqrt{y^2+3}+y > |y|+y \geq 0[/TEX]
[TEX]gt \Leftrightarrow ( \sqrt{x^2+3}+x)( \sqrt{y^2+3}+y)=( \sqrt{y^2+3}+y)( \sqrt{y^2+3}-y)[/TEX]
[TEX]\Leftrightarrow \sqrt{x^2+3}+x= \sqrt{y^2+3}-y[/TEX]
[TEX]\Leftrightarrow (x+y)+( \sqrt{x^2+3}- \sqrt{y^2+3})=0[/TEX]
[TEX]\Leftrightarrow (x+y)+ \frac{x^2-y^2}{ \sqrt{x^2+3}+ \sqrt{y^2+3}}=0[/TEX]
[TEX]\Leftrightarrow x+y=0(1),hoac: 1+ \frac{x-y}{ \sqrt{x^2+3}+ \sqrt{y^2+3}}=0(2)[/TEX]
Do [TEX] \sqrt{x^2+3}+ \sqrt{y^2+3} > |x|+|y| \geq |x-y|[/TEX]
[TEX]\Rightarrow \frac{|x-y|}{ \sqrt{x^2+3}+ \sqrt{y^2+3}} < 1[/TEX]
[TEX]\Rightarrow 1+ \frac{|x-y|}{ \sqrt{x^2+3}+ \sqrt{y^2+3}} > 0 \Rightarrow (2) VN[/TEX]
Vậy x+y=0 \Leftrightarrow y=-x
[TEX]\Rightarrow A=x^{2009}+(-x)^{2009}+1=x^{2009}-x^{2009}+1=1[/TEX]
 
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thaopro1230

Ta có [TEX]p^2[/TEX]
=[TEX](x+\sqrt{x}+1)^2[/TEX]
=[TEX]x^2+x+1+2x\sqrt{x}+2x+2\sqrt{x}[/TEX]
Mà 2p=[TEX]2x+2\sqrt{x}+1[/TEX]
Vì [TEX]x\geq0[/TEX][TEX]\sqrt{x}\geq0[/TEX]
[TEX]x^2+x+1+2x\sqrt{x}+2x+2\sqrt{x}\geq2x+2\sqrt{x}+1[/TEX]
[TEX]\Rightarrow p^2\geq2p[/TEX]
 
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khanhtoan_qb

mamy007 said:
2)cho 2 số x,y thỏa mãn ([TEX]\sqrt{x^2+3}+x[/TEX])([TEX]\sqrt{y^2+3}+y[/TEX]=3
tính A=[TEX] x^{2009} + y^{2009} +1[/TEX]
Bấm để xem đầy đủ nội dung ...
Bài 2:
ta có:
[TEX]\sqrt{x^2+3}+x)(\sqrt{y^2+3}+y = (\sqrt{x^2 + 3} + x)(\sqrt{x^2 + 3} - x) = 3[/TEX]
\Rightarrow [TEX]\sqrt{y^2 + 3} + y = \sqrt{x^2 + 3} - x[/TEX]
T ương tự có:
[TEX]\sqrt{y^2 + 3} - y = \sqrt{x^2 + 3} + x[/TEX]
Công lại được:
[TEX]x + y = - (x + y) \Rightarrow x + y = 0 \Rightarrow A = 1[/TEX]
 
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