lam on cho e hoi nay voi ak!..

N

nguyenbahiep1

vi sao tu [ sqrt{1+x}-sqrt{1-x}]^2nho hon hoac=x^2 bien doi thanh (1-sqrt{1-x})^2 nho hon hoac=0 duoc ak? giup e voi ak! e cam on nhieu!


[laTEX](\sqrt{1+x}-\sqrt{1-x})^2 = 1 + x - 2.\sqrt{1+x}.\sqrt{1-x} + 1-x = 2 -2.\sqrt{1-x^2}[/laTEX]

vậy

[laTEX](\sqrt{1+x}-\sqrt{1-x})^2 \leq x^2 \Leftrightarrow 2 -2\sqrt{1-x^2} \leq x^2 \\ 2 -x^2 - 2.\sqrt{1-x^2} \leq 0 \\ 1 -x^2 - 2.\sqrt{1-x^2} + 1 \leq 0 \Leftrightarrow ( 1- \sqrt{1-x^2})^2 \leq 0 [/laTEX]

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