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9.3: Ongeza na Ondoa Maneno ya busara

  • Page ID
    164730
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    Ufafanuzi: Ongeza au Ondoa Maneno ya busara

    Ili kuongeza au kuondoa maneno ya busara, fikiria hili kama sehemu ndogo na vigezo. Denominator ya kawaida (inayoitwa LCD) inahitajika kwa kuongeza na kuondoa.

    Pata LCD/LCM

    Ufafanuzi: LCD/LCM

    Ili kupata LCD, kwanza factor kikamilifu denominators wote. Kujenga LCD kutoka kwa sababu zilizopatikana katika denominators zote. Kuzidisha kila sababu idadi kubwa ya nyakati hutokea katika kujieleza ama. Ikiwa sababu hiyo hutokea zaidi ya mara moja katika maneno yote mawili, kuzidisha sababu idadi kubwa ya nyakati hutokea katika kujieleza ama. Itaitwa LCM katika sehemu hii (Multiple Multiple Common), kwa sababu hakuna sehemu ndogo katika matatizo haya.

    1. \((x^2 − 2x − 3)\)na\((x^2 + 2x − 15)\)
    2. \((x^2 − 9)\)na\((2x^2 − 5x − 3)\)
    3. \((x^2 + x − 2)\)na\((x^2 + 4x + 4)\)

    Suluhisho

    1. \(\begin{array} &&(x^2 − 2x − 3) \text{ and } (x^2 + 2x − 15) &\text{Example problem} \\ &(x − 3)(x + 1) \text{ and } (x − 3)(x + 5) &\text{Factor} \\ &\text{The LCM is } (x − 3)(x + 1)(x + 5) &\text{Final answer. Every factor must be represented in the answer.} \\& &\text{Only one copy of \((x − 3)\)inahitajika, kwa sababu inawakilisha sababu inayopatikana katika kila kujieleza.} \ mwisho {safu}\)
    1. \(\begin{array} &&(x^2 − 9) \text{ and } (2x^2 − 5x − 3) &\text{Example problem} \\ &(x−3)(x+3) \text{ and } (2x^2−6x+1x−3) &\text{Factor; the first polynomial is a difference of squares, and use factor by grouping for the second polynomial.} \\ &(x−3)(x+3) \text{ and } (2x(x−3)+1(x− 3)) &\text{Factor by grouping.} \\ &(x − 3)(x + 3) \text{ and } (2x + 1)(x − 3) &\text{Completely factored.} \\ &\text{The LCM is } (x − 3)(x + 3)(2x + 1) &\text{Final answer. Every factor must be represented in the answer.} \\ & &\text{Only one copy of \((x − 3)\)inahitajika, kwa sababu inawakilisha sababu inayopatikana katika kila kujieleza.} \ mwisho {safu}\)
    1. \(\begin{array} &&(x^2 + x − 2) \text{ and } (x^2 + 4x + 4) &\text{Example problem} \\ &(x − 1)(x + 2) \text{ and } (x + 2)(x + 2) &\text{Factor.} \\ &\text{The LCM is } (x − 1)(x + 2)(x + 2) &\text{Final answer. Every factor must be represented in the answer.} \\ & &\text{Two copies of \((x + 2)\)zinahitajika, kwa sababu inawakilisha idadi kubwa ya mambo hayo yanayopatikana katika kujieleza ama.}\\ &\ maandishi {LCM ni} (x - 1) (x + 2) ^2 &\ maandishi {Jibu mbadala.} \ mwisho {safu}\)

    Pata LCM:

    1. \((3x^2 − 13x + 4)\)na\((x^2 − 16)\)
    2. \((2x^2 + x − 3)\)na\((x^2 − 2x + 1)\)
    3. \((x − 1)\)na\((x^2 − 4x − 5)\)
    4. \((6x^2 − 23x + 20)\)na\((4x^2 − 25)\)

    Ondoa Maneno ya busara na Urahisishe kwa kujieleza moja kwa moja

    Ufafanuzi: Ongeza au Ondoa Maneno ya busara Kutumia LCD

    Maneno ya busara ni sehemu ndogo na vigezo (pia inajulikana kama sehemu za algebraic). Ili kuongeza au kuondoa maneno ya busara, kwanza pata denominator ya kawaida (LCD), kisha uongeze au uondoe namba, ukiweka denominator sawa (ya kawaida). Hatimaye, sababu na kurahisisha kwa kuondoa mambo ya kawaida kutoka kwa nambari na denominator ikiwezekana.

    Kuongeza au Ondoa na kurahisisha:

    1. \(\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5}\)
    2. \(\dfrac{4}{x^2 − 9} - \dfrac{5}{x^2 − 6x + 9}\)
    3. \(\dfrac{x}{1 + x} + \dfrac{2x + 3}{x^2 − 1}\)

    Suluhisho

    1. \(\begin{array} &&\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5} &\text{Example problem} \\ &\dfrac{2x}{2x − 1} - \dfrac{2x}{2x + 5} &\text{Find the LCD, which is \((2x − 1)(2x + 5)\)}\\ &\ dfrac {2x (2x + 5)} {(2x - 1) (2x + 5)} -\ dfrac {2x (2x - 1)} {(2x - 1) (2x + 5)} &\ Nakala {Kuzidisha nambari na denominator ya kila kujieleza kwa busara kwa maneno yaliyopo kwenye LCD.}\\ &\ DFRAC {2x (2x + 5) - [2x (2x - 1)]} {(2x - 1) (2x + 5)} &\ maandishi {Weka uondoaji katika nambari juu ya denominator moja ya kawaida.}\\ &\ dfrac {4x^2 + 10x - [4x^2 - 2x]} {(2x - 1) (2x + 5)} &\ maandishi {Kusambaza, kuchanganya maneno kama na kurahisisha namba.}\\ &\ dfrac {4x^2 + 10x ≈ 4x^2 + 2x} {(2x ≈ 1) {(2x ≈ 1) (2x + 5)} &\ maandishi {Jihadharini na kusambaza uondoaji kwa maneno yote mawili.}\\ &\ dfrac { 12x} {(2x - 1) (2x + 5)} &\ maandishi {Jibu la mwisho.} \ mwisho {safu}\)
    1. \(\begin{array} &&\dfrac{4}{x^2 − 9} - \dfrac{5}{x^2 − 6x + 9} &\text{Example problem} \\ &\dfrac{4}{(x + 3)(x − 3)} - \dfrac{5}{(x − 3)(x − 3)} &\text{Factor the denominators.} \\ &\dfrac{4}{(x + 3)(x − 3)} - \dfrac{5}{(x − 3)(x − 3)} &\text{Find the LCD, which is \((x − 3)(x − 3)(x + 3)\)}\\ &\ dfrac {4 (x - 3)} {(x + 3) (x -3) (x -3)} -\ dfrac {5 (x + 3)} {(x - 3) (x -3) (x -3) (x + 3)} &\ maandishi {Panua namba na denominator ya kila kujieleza kwa maneno yaliyopo katika LCD.}\\ &\ dfrafrc c {4} {(x - 3) - 5 (x + 3)} {(x + 3) (x - 3) (x - 3)} &\ maandishi {Weka uondoaji nambari juu ya denominator moja ya kawaida.}\\ &\ dfrac {4x - 12 - [5x + 15]} {(x + 3) (x -3) (x - 3)} &\ maandishi {Kusambaza, kuchanganya maneno kama na kurahisisha namba.}\\ &\ dfrac {4x ÷ 12 ≈ 5x ≈ 15} {(x + 3) (x ≈ 3) (x ≈ 3) (x - 3)} &\ Nakala {Jihadharini kusambaza uondoaji kwa maneno yote mawili.}\\ &\ dfrac {-x - 27} {(x + 3) (x -3) (x -3)} &\ maandishi {Jibu la mwisho.}\\ &\ dfrac {(x + 27)} {(x + 3) (x + 3) (x - 3) (x - 3)} &\ maandishi {Jibu mbadala}\ mwisho {safu}\)
    1. \(\begin{array} &&\dfrac{x}{1 + x} + \dfrac{2x + 3}{x^2 − 1} &\text{Example problem} \\ &\dfrac{x}{x + 1} + \dfrac{2x + 3}{(x − 1)(x + 1)} &\text{Factor the denominators.} \\ &\dfrac{x}{x + 1} + \dfrac{2x + 3}{(x − 1)(x + 1)} &\text{Find the LCD, which is \((x − 1)(x + 1)\)}\\ &\ dfrac {x (x - 1)} {(x + 1) (x - 1)} +\ dfrac {(2x + 3)} {(x - 1) (x + 1)} &\ maandishi {Kuzidisha nambari na denominator ya kila kujieleza kwa busara kwa maneno yaliyopo kwenye LCD.}\\ & & &\ maandishi {Angalia kuwa kujieleza kwa busara ya pili tayari ina LCD kama denominator yake.}\\ [0.125 katika] &\ dfrac {x (x - 1) + (2x + 3)} {(x + 1) (x - 1)} &\ maandishi {Weka uondoaji katika nambari juu ya denominator moja ya kawaida.}\\ &\ dfrac {x^2 - x + 2x + 3} {(x + 1) (x - 1)} &\ maandishi {Kusambaza, kuchanganya kama masharti na kurahisisha tarakimu.}\\ &\ dfrac {x ^ 2 + x + 3} {(x + 1) (x - 1)} &\ maandishi { Jihadharini kusambaza uondoaji kwa maneno yote mawili.}\\ &\ dfrac {x ^ 2 + x + 3} {(x + 1) (x - 1)} &\ maandishi {Jibu la mwisho.} \ mwisho {safu}\)

    Kuongeza au Ondoa na kurahisisha:

    1. \(\dfrac{x}{x^2 + 1} + \dfrac{24x^3}{x3 + 2}\)
    2. \(\dfrac{x}{1 − x} + \dfrac{2x + 3}{x^2 − 1}\)
    3. \(\dfrac{5}{x + 3} + \dfrac{x^2 − 4x − 21}{x^2 − 9}\)
    4. \(\dfrac{39x + 36}{x^2 − 3x − 10} - \dfrac{23x − 16}{x^2 − 7x + 10}\)