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Matemātika I 11 (2024./2025.m.g.)

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Atrisini doto nevienādību, izmantojot intervālu metodi! (5p - 1p par katru būtisko darbību)

fraction numerator x minus 1 over denominator 2 end fraction plus fraction numerator 3 over denominator x plus 2 end fraction less or equal than fraction numerator 2 x plus 3 over denominator 4 end fraction plus fraction numerator x over denominator x plus 2 end fraction

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Kuras nevienādības atrisinājums atbilst zīmējumam?            

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begin mathsize 14px style N e v i e n ā d ī b a s space space fraction numerator 3 space minus space 2 x over denominator 3 space minus space x end fraction space greater than space 4 space space space a t r i sin ā j u m u space k o p a space i r space.... end style

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Nevienādības  ( x - 5 )•( x + 3 ) < 0  atrisinājumu kopa ir....

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begin mathsize 14px style space N e v i e n ā d ī b a s space space space fraction numerator x space plus space 2 over denominator 3 space minus space x end fraction space space less than space 0 space space a t r i sin ā j u m a space k o p a space i r... end style

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begin mathsize 14px style space N e v i e n ā d ī b a s space space space fraction numerator 4 space minus space x over denominator x to the power of 2 space end exponent minus space 4 end fraction space greater than space 0 space a t r i sin ā j u m a space k o p a space i r space... end style

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begin mathsize 14px style N e v i e n ā d ī b a s space space space space space fraction numerator x space minus space 5 over denominator x space minus space 1 end fraction less than space 0 space space space space a t r i sin ā j u m a space k o p a space i r space.... end style

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begin mathsize 14px style space V a i space n e v i e n ā d ī b a s space space space fraction numerator x space plus 1 over denominator x end fraction greater than space 0 space space space space space u n space space space x left parenthesis space x space plus space 1 space right parenthesis space greater than space 0 space space i r space e k v i v a l e n t a s ? end style

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Atrisini doto vienādojumu, parādot pilnu aprēķina gaitu!

fraction numerator x over denominator x minus 2 end fraction minus fraction numerator 7 over denominator x plus 2 end fraction equals fraction numerator 8 over denominator x squared minus 4 end fraction

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Nosaki dotā vienādojuma atrisinājumu!

fraction numerator x plus 3 over denominator 5 end fraction minus x over 2 equals fraction numerator 3 x over denominator 10 end fraction

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