Skip to main content
Query

9.1: Kurahisisha maneno ya busara

  • Page ID
    164731
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Ufafanuzi: Maneno ya busara

    Maneno ya busara yameandikwa kama quotient ya polynomials

    \[\dfrac{P(x)}{Q(x)} \nonumber \]

    wapi\(P(x)\) na\(Q(x)\) ni polynomials katika variable moja\(x\).

    Ili kurahisisha kujieleza kwa busara, fikiria namba zote na denominator, na uondoe mambo ya kawaida kutoka kwa namba zote na denominator. Maneno rahisi ya busara yana mgawanyiko mmoja tu, na namba moja na denominator. Ikiwa maneno hayawezi kuhesabiwa, basi usemi wa busara hauwezi kurahisishwa.

    Kurahisisha maneno ya busara:

    1. \(\dfrac{x^2 + 2x − 3}{x^2 + 4x + 3}\)
    2. \(\dfrac{(x^2 + 1)^2 (−2) + (2x)(2)(x^2 + 1)(2x)}{(x^2 + 1)^4}\)
    3. \(\dfrac{(x^2 + 1) \frac{1}{2} (x^{−\frac{1}{2}}) − (2x)(x^{\frac{1}{2}})}{(x^2 + 1)^2}\)

    Suluhisho

    1. \(\begin{array} &&\dfrac{x^2 + 2x − 3}{x^2 + 4x + 3} &\text{Example problem} \\ &\dfrac{(x + 3)(x − 1)}{(x + 3)(x + 1)} &\text{Factor both numerator and denominator.} \\ &\dfrac{\cancel{(x + 3)}(x − 1)}{\cancel{(x + 3)}(x + 1)} &\text{Remove common factors, because \(\dfrac{x + 3}{x + 3} = 1\)}\\ &\ dfrac {x - 1} {x + 1} &\ maandishi {jibu la mwisho}\ mwisho {safu}\)
    1. \(\begin{array} &&\dfrac{(x^2 + 1)^2 (−2) + (2x)(2)(x^2 + 1)(2x)}{(x^2 + 1)^4} &\text{Example problem} \\ &\dfrac{2(x^2 + 1)[(x^2 + 1)(−1) + (2x)(2x)]}{(x^2 + 1)^4} &\text{Factor out 2(x^2 + 1)} \\ &\dfrac{2 \cancel{(x^2 + 1)}[(x^2 + 1)(−1) + (2x)(2x)]}{\cancel{(x^2+1)}(x^2 + 1)^3} &\text{Remove common factors, because \(\dfrac{x^2 + 1}{x^2 + 1} = 1\)}\\ &\ dfrac {2 [-x^2 - 1 + 4x^2]} {(x ^ 2 + 1) ^3} &\ maandishi {Kurahisisha kwa kuzidisha na kuchanganya kama maneno}\\ &\ dfrac {2 (3x^2 - 1)} {(x ^ 2 + 1) ^3} &\ maandishi {Jibu la mwisho}\ mwisho {safu}\)
    1. \(\begin{array} &&\dfrac{(x^2 + 1) \frac{1}{2} (x^{−\frac{1}{2}}) − (2x)(x^{\frac{1}{2}})}{(x^2 + 1)^2} &\text{Example problem} \\ &\dfrac{\frac{(x^2+1)}{2x^{\frac{1}{2}}} − (2x)(x^{ \frac{1}{2} })}{(x^2 + 1)^2} &\text{Work with the negative exponent in the first term of the numerator by moving the factor to the denominator of the first term, next to the \(2\).}\\ &\ dfrac {(x ^ 2 + 1) - (2x) (x^ {\ Frac {1} {2}}) 2 (x^ {\ Frac {1} {2}}}} {\ dfrac {2x^ {1} {2}} {2}} {(x ^ 2 + 1) ^2}} &\ maandishi {dhehebu ya kawaida ator}\\ &\ dfrac {x ^ 2 + 1 - 4x^ 2} {(2x^ {\ Frac {1} {2}}) (x ^ 2 + 1) ^2} &\ maandishi {Kurahisisha kwa kuzidisha na kuchanganya maneno kama}\\ &\ dfrac {-3x^2 + 1} {(2x^ {\ frac {1} {2}}) (x ^ 2 + 1) ^2} &\ maandishi {jibu la mwisho}\ mwisho {safu}\)

    Kurahisisha maneno ya busara:

    1. \(\dfrac{2x^2 + 3x − 2}{2x^2 + 5x − 3}\)
    2. \(\dfrac{(t^2 + 4)(2t − 4) − (t^2 − 4t + 4)(2t)}{(t^2 + 4)^2}\)
    3. \(\dfrac{(2)(x − 4)(x^2 + 4x + 4)}{(x + 2)(x^2 − 16)}\)
    4. \(\dfrac{12x^2 + 19x − 21}{12x^2 + 38x − 40}\)