1) √(2x+1+x^2)+√(x-2√(1+x)+2) при x=-0,91 2) 3x+√(4x^2+4x+1) при x=-0,6 3) x+√(x^2-6x+9) при x=2,007. Created by folium8780. algebra-ru

1) √(2x+1+x^2)+√(x-2√(1+x)+2) при x=-0,91 2) 3x+√(4x^2+4x+1) при x=-0,6 3) x+√(x...

folium8780 @folium8780

July 2022 1 10 Report

1) √(2x+1+x^2)+√(x-2√(1+x)+2) при x=-0,91 2) 3x+√(4x^2+4x+1) при x=-0,6 3) x+√(x^2-6x+9) при x=2,007

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Dимасuk

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$1) \sqrt{2x + 1 + x^2} + \sqrt{x - 2 \sqrt{1 + x} + 2 } = \sqrt{(x + 1)^2}+ \\ + \sqrt{x + 1 - 2 \sqrt{1 + x} + 1} = |x + 1| + \sqrt{( \sqrt{x + 1} - 1 )^2} = |x + 1| + \\ + |\sqrt{x + 1} - 1| = |1 - 0,91| + | \sqrt{1 - 0,91} - 1| = 0,09 + |0,3 - 1| = \\ = 0,09 + 0,7 = 0,79$

$2) \ 3x + \sqrt{4x^2 + 4x + 1} = 3x + \sqrt{(2x + 1)^2} = 3x + |2x + 1| = 3 = \\ = 3 \cdot (-0,6) + |2 \cdot (-0,6)+ 1| = -1,8 + |-0,2| = -1,8 + 0,2 = -1,6 
\\ \\ 3) \ x + \sqrt{x^2 - 6x + 9} = x + \sqrt{(x - 3)^2} = x + |x - 3| = 2,007 + \\ + |2,007 - 3| = 2,007 - 2,007 + 3 = 3$

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folium8780 July 2022 | 0 Ответы

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