Toán 9 Tìm GTLN của P

Minh Dora said:

[tex]=(\frac{\sqrt{x-2}}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}).\frac{(1-x)^2}{2}[/tex]
[tex]=\frac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\frac{(1-x)^2}{2} =\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\sqrt{x}+1}.\frac{-1}{2} =\frac{\sqrt{x}}{\sqrt{x}+1}=1-\frac{1}{\sqrt{x}+1}\leq 1-\frac{1}{1}=0[/tex]
Dấu = xảy ra khi căn x=0 suy ra x=0

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Minh Dora said:

[tex]=(\frac{\sqrt{x-2}}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}).\frac{(1-x)^2}{2}[/tex]
[tex]=\frac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\frac{(1-x)^2}{2} =\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\sqrt{x}+1}.\frac{-1}{2} =\frac{\sqrt{x}}{\sqrt{x}+1}=1-\frac{1}{\sqrt{x}+1}\leq 1-\frac{1}{1}=0[/tex]
Dấu = xảy ra khi căn x=0 suy ra x=0

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Minh Dora said:

[tex]=(\frac{\sqrt{x-2}}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}).\frac{(1-x)^2}{2} =\frac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\frac{(1-x)^2}{2} =\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{+1}.\frac{\sqrt{x}-1}{2} =\sqrt{x}-x =\frac{1}{4}-(x-\sqrt{x}+\frac{1}{4})=\frac{1}{4}-(x-\frac{1}{2})^2 <= \frac{1}{4}[/tex]
Dấu = xảy ra<=> x=1/2

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