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5.4: Utawala wa sifuri

  • Page ID
    164584
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    Katika kifungu cha 5.3, kielelezo cha nambari katika namba kilikuwa kikubwa zaidi kuliko kielelezo cha idadi katika denominator.

    Katika kifungu cha 5.4, kielelezo cha nambari katika nambari kitakuwa sawa na kielelezo cha idadi katika denominator.

    Ufafanuzi: Utawala wa Ufafanuzi wa Zero

    Kwa idadi yoyote halisi\(a\), Utawala wa Zero Exponent ni yafuatayo

    \(a^0= 1\)

    Wazo:

    Kutoka sehemu zilizopita:

    \[x^5 = x \cdot x \cdot x \cdot x \cdot x \nonumber \]

    na

    \[\dfrac{x^5 }{x^5} =\dfrac{ x \cdot x \cdot x \cdot x \cdot x }{x \cdot x \cdot x \cdot x \cdot x }= \dfrac{\cancel{x \cdot x \cdot x \cdot x \cdot x }}{\cancel{x \cdot x \cdot x \cdot x \cdot x }}= 1 \nonumber \]

    Hivyo,

    \[\dfrac{x ^5 }{x^5} = x^{5−5 }= x^0=1 \nonumber \]

    Tumia utawala wa sifuri ili kurahisisha maneno.

    1. \(\dfrac{x^9 }{x^9}\)
    2. \(\dfrac{d^5 }{d^2 \cdot d^3}\)
    3. \(\dfrac{5(xy)^3 }{(xy)^3}\)
    4. \(-\dfrac{y^3 }{\sqrt{5}y^3}\)
    5. \(\dfrac{(ab^2 )^7 }{(ab^2)^2 \cdot (ab^2)^4 ˙(ab^2)}\)
    Suluhisho
    Ufafanuzi Zero exponent Kanuni
    \(\dfrac{x^9 }{x^9}\) \(x^{9−9} = x^0 = 1\)
    \(\dfrac{d^5 }{d^2 \cdot d^3}\) \(\dfrac{d^5 }{d^{2+3 }}= \dfrac{d^5 }{d^5} = d^{5−5 }= d^{0} = 1\)
    \(\dfrac{5(xy)^3 }{(xy)^3}\)

    \(5 \cdot \dfrac{(xy)^3 }{(xy)^3 }= 5 \cdot (xy)^{3−3 }= 5 \cdot (xy)^0 = 5 \cdot 1 = 5 \)

    5 mara kwa mara, inaweza kuwa factored nje ya kuona besi ya kawaida wazi.

    \(-\dfrac{y^3 }{\sqrt{5}y^3}\)

    \(− \dfrac{1}{ \sqrt{5}} \cdot \dfrac{y^3 }{y^3 }= − \dfrac{1}{ \sqrt{5}} \cdot y ^{3−3 }= − \dfrac{1}{ \sqrt{5}} \cdot y^0 = − \dfrac{1}{ \sqrt{5}} \cdot 1 = − \dfrac{1}{ \sqrt{5}}\)

    Mara kwa mara\(−\left( \dfrac{1 }{\sqrt{5}}\right )\), inaweza kuzingatiwa ili kuona misingi ya kawaida wazi.

    \(\dfrac{(ab^2 )^7 }{(ab^2)^2 \cdot (ab^2)^4 ˙(ab^2)}\)

    \(\dfrac{(ab^2 )^7 }{(ab^2)^{2+4+1}}= \dfrac{(ab^2 )^7}{ (ab^2)^7} = (ab^2 )^{7−7 }= (ab^2 )^0 = 1\)

    Kwanza, kurahisisha denominator kutumia utawala wa bidhaa wa exponents. Kisha kutumia utawala quotient ya exponents kurahisisha kujieleza iliyobaki.

    Kumbuka:\(0^0\) si sawa 1. Hii ni kesi maalum ambayo inafunikwa katika kozi za juu. Kwa sasa fikiria\(0^0\) kuwa haijulikani.

    Hatua muhimu ili kurahisisha maneno na vielelezo

    1. Tambua misingi ya kawaida.
    2. Ikiwa inahitajika kuchanganya besi za kawaida kwa kutumia utawala wa bidhaa wa exponents.
    3. Ikiwa maneno yana misingi ya kawaida katika namba zote mbili na denominator, tumia utawala wa quotient wa vielelezo kama inahitajika.

    Tumia sheria zote za exponents kufunikwa hadi sasa katika sura hii ili kurahisisha zifuatazo.

    1. \(\dfrac{z ^4 }{z^ 4}\)
    2. \(\dfrac{d^2 \cdot d^8}{ d^7 \cdot d^3}\)
    3. \(\dfrac{5(x + y)^3 }{2(x + y)^3}\)
    4. \(−\dfrac{\sqrt{9}{y^3 }}{y^3}\)
    5. \(\dfrac{(a^3b^2 )^9}{ (a^3b^2)^3 \cdot (a^3b^2)^4 ˙(a^3b^2)^2}\)
    6. \(\dfrac{(xyz)^{19} }{(xyz)^{19}}\)