решить неравенство [tex] x^{2} [/tex][tex] \leq [/tex]1+2x. Created by spodyninogin. matematika-ru

tex]1+2x...

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решить неравенство [tex] x^{2} [/tex][tex] \leq [/tex]1+2x

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ТатМих

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$x^{2} -2x-1 \leq 0 \\ D=4+4=8 \\ \sqrt{D} = \sqrt{8} =2 \sqrt{2} \\ x1= \frac{2+2 \sqrt{2} }{2} =1+ \sqrt{2} \\ \\ x2=1- \sqrt{2}$

+ - +
........1-√2...........1+√2........

Ответ:х=[1-√2;1+√2].

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reshite 4x 13gt3x 10 11 4xlt12 3x

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