toan ptmu (giai cap toc!)

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0samabinladen

codocvuog said:
moi nguoi oi giai dum toi bai nay voi
cho 3 so thuc x, y, z sao cho x+y+z=0
tim Minh cua P= [tex]\sqrt{2^x + 3} + \sqrt{2^y + 3} + \sqrt{2^z + 3}[/tex]
thanhk moi nguoi nhieu lam!!!!!!!!!!!!
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đặt [TEX]vecto u=(\sqrt{2^x};\sqrt{3}); v=(\sqrt{2^y};\sqrt{3}); p=(\sqrt{2^z};\sqrt{3})[/TEX]

[TEX]Theo BDT vecto[/TEX]

[TEX]|vt.u|+|vt.v|+|vt.p|\geq |vt.u+vt.v+vt.p|[/TEX]

[TEX]\leftrightarrow P=\sqrt{2^x + 3} + \sqrt{2^y + 3} + \sqrt{2^z + 3}\geq\sqrt{(\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z})^2+27}[/TEX]

[TEX]BDT.cosi [/TEX]

[TEX]\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z} \geq 3\sqrt[3]{\sqrt{2^{(x+y+z)}}}=3[/TEX]

[TEX](\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z})^2 \geq 9[/TEX]

[TEX]\longrightarrow P \geq 6[/TEX]

[TEX]\longrightarrow P_{min}=6 \leftrightarrow x=y=z=0[/TEX]
 
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S

son5c

0samabinladen said:
đặt [TEX]vecto u=(\sqrt{2^x};\sqrt{3}); v=(\sqrt{2^y};\sqrt{3}); p=(\sqrt{2^z};\sqrt{3})[/TEX]

[TEX]Theo BDT vecto[/TEX]

[TEX]|vt.u|+|vt.v|+|vt.p|\geq |vt.u+vt.v+vt.p|[/TEX]

[TEX]\leftrightarrow P=\sqrt{2^x + 3} + \sqrt{2^y + 3} + \sqrt{2^z + 3}\geq\sqrt{(\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z})^2+27}[/TEX]

[TEX]BDT.cosi [/TEX]

[TEX]\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z} \geq 3\sqrt[3]{\sqrt{2^{(x+y+z)}}}=3[/TEX]

[TEX](\sqrt{2^x}+\sqrt{2^y}+\sqrt{2^z})^2 \geq 9[/TEX]

[TEX]\longrightarrow P \geq 6[/TEX]

[TEX]\longrightarrow P_{min}=6 \leftrightarrow x=y=z=0[/TEX]
Bấm để xem đầy đủ nội dung ...
Dùng AM-GM:
Đặt [TEX]2^x=a[/TEX] [TEX]2^y=b[/TEX] [TEX]2^z=c[/TEX] ta có: [TEX]abc=2^(x+y+z)=1[/TEX]
[TEX]P\geq3\sqrt[3]{\sqrt{(a+3)(b+3)(c+3)}[/TEX]
[TEX]a+3=a+1+1+1\geq4\sqrt[4]{a}[/TEX]
.................................................................
.................................................................
[TEX] \Rightarrow (a+3)(b+3)(c+3)\geq64\sqrt[4]{abc}\geq64[/TEX]
[TEX] \Rightarrow P \geq 6[/TEX]
 
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