// Numbas version: finer_feedback_settings {"name": "Irrationale Zahlen - Nenner rational machen", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Irrationale Zahlen - Nenner rational machen", "tags": [], "metadata": {"description": "
Rationalise the denominator with increasingly difficult examples involving compound denominators.
\nTranslated to German, made some minor changes to Advice section.
\nOriginal: https://numbas.mathcentre.ac.uk/question/22555/rationalising-the-denominator-surds/ by Lauren Richards
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Schreiben Sie die folgenden Zahlen als Brüche mit einer ganzen Zahl im Nenner.
", "advice": "Es gilt $\\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\frac{a}{b}}$.
\ni)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{{\\var{prime_nums[0]}}}} = \\sqrt{\\var{square_nums[0]}}= \\simplify{{sqrt(square_nums[0])}}$
\nii)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{{\\var{prime_nums[1]}}}} = \\sqrt{\\var{square_nums[1]}} = \\simplify{{sqrt(square_nums[1])}}$.
\nUm eine ganze Zahl in den Nenner zu bekommen, benutzen wir, dass $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$.
\nGefragt ist, den Ausdruck $\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}$ umzuschreiben als Bruch mit einer ganzen Zahl im Nenner.
\nDer Nenner ist ${\\sqrt{\\var{m_lcm}}}$, also multiplizieren wir den Nenner mit ${\\sqrt{\\var{m_lcm}}}$.
\n\\[ \\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}\\cdot\\frac{\\sqrt{\\var{m_lcm}}}{\\sqrt{\\var{m_lcm}}} \\]
\nWeil $\\sqrt{a}\\cdot\\sqrt{b}$ = $\\sqrt{ab}$, können wir den Zähler als $\\sqrt{\\var{num}}$ schreiben.
\nDer Nenner ist ${\\sqrt{\\var{m_lcm}}}\\cdot{\\sqrt{\\var{m_lcm}}} = {\\var{m_lcm}}$.
\nAlso ist das Ergebnis dieser Rechnung
\n\\[ \\frac{\\sqrt{\\var{num}}}{\\var{m_lcm}} \\]
\nZum Schluss können wir noch kürzen und erhalten
\n\\[ \\frac{\\sqrt{\\var{a[0]*b[0]}}}{\\var{b[0]}} \\]
\n(Man kann natürlich auch schon zu Beginn kürzen, das ist im Grunde effizienter.)
\nZunächst erweitern wir den Bruch mit $\\sqrt{\\var{f}}$. Weil $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$ gilt, ist der Nenner des entstehenden Bruchs $\\var{d}\\cdot\\var{f}=\\var{df}$.
\n$\\displaystyle\\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}}\\cdot\\simplify[all,!simplifyFractions,!sqrtDivision]{sqrt({f})/(sqrt({f}))}={\\frac{\\simplify{{c}*sqrt({f})}}{{(\\var{d}\\cdot\\var{f})}}}=\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}$.
\nWir kürzen noch mit $\\var{gcd_cdf}$, und erhalten als Ergebnis
\n\\[\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}=\\frac{\\simplify{{c_coprime}*sqrt({f})}}{\\var{df_coprime}}\\]
\nDie Antwort soll in der Form $\\frac{a \\sqrt{\\var{f}}}{b}$ gegeben werden, also ist $a = \\var{c_coprime}$ und $b = \\var{df_coprime}$.
\n\nUm auf eine ganze Zahl im Nenner zu kommen, wenn der Nenner eine Summe ist wie bei $\\displaystyle\\frac{\\var{g}}{\\sqrt{\\var{h}}+\\var{j}}$, erweitern wir mit $\\sqrt{\\var{h}}-\\var{j}$ und benutzen die dritte binomische Formel.
\n\\[ \\frac{\\var{g}}{\\sqrt{\\var{h}}+\\var{j}}\\cdot\\frac{(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}-\\var{j})}=\\frac{\\var{g}(\\sqrt{\\var{h}}-\\var{j})}{(\\sqrt{\\var{h}}+\\var{j})(\\sqrt{\\var{h}}-\\var{j})}\\text{.} \\]
\nWir erhalten
\n\\[ \\frac{ \\simplify{ {g*-j} + {g}*sqrt({h})} }{\\var{h-j^2}}\\text{,} \\]
\nDas können wir noch vereinfachen zu
\n\\[ \\var{o}\\sqrt{\\var{h}}-{\\var{l}}\\text{.} \\]
\n\nWir rechnen
\n$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}$ =
\n$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}\\times\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}-\\sqrt{\\var{u}}}=\\frac{(\\var{t}-\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}{(\\var{t}+\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}=\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}$.
\n$\\displaystyle\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}=\\frac{\\var{t^2+u}-\\var{2t}{\\sqrt{\\var{u}}}}{\\var{t^2-u}}$.
\nKürzen mit ${\\var{gcd_frac}}$ ergibt dann
\n$\\displaystyle\\frac{\\var{num_simp_1}-\\simplify{{num_simp_2}*sqrt({u})}}{\\var{denom_simp}}$.
\nUm die Brüche $\\displaystyle\\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$ zu addieren, müssen wir sie zunächst auf einen gemeinsamen Nenner bringen.
\nBeachten Sie, dass $\\sqrt{\\var{p^3}}$ umgeschrieben werden kann als $\\var{p}\\sqrt{\\var{p}}$, das bedeutet, dass wir den ersten Bruch schreiben können als
\n\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\var{p}\\sqrt{\\var{p}}} \\text{.} \\]
\nDas vereinfachen wir weiter zu
\n\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}} \\text{.} \\]
\nDeshalb haben wir als Zwischenergebnis $\\displaystyle\\frac{\\var{p}}{{\\var{p+1}}\\sqrt{\\var{p}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$.
\nUm die Addition durchzuführen, müssen wir den zweiten Bruch auf denselben Nenner bringen. Wir erweitern mit $\\simplify{{p_p1}sqrt({p})}$, also
\n$\\displaystyle\\frac{\\var{r_coprime}}{\\var{s_coprime}}\\cdot\\simplify[all,!simplifyFractions,!sqrtDivision]{{p_p1}*sqrt({p})/({p_p1}*sqrt({p}))}=\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}$,
\nJetzt erhalten wir
\n\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}}+\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}= \\frac{\\var{p}+\\simplify{{one}*sqrt({p})}}{\\var{p+1}\\sqrt{\\var{p}}} \\]
\nWeil $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$ ist, können wir $\\sqrt{\\var{p}}$ im Zähler ausklammern:
\n\\[ \\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}\\text{.} \\]
\nWir kürzen jetzt mit $\\sqrt{\\var{p}}$ und bekommen
\n\\[ \\frac{\\sqrt{\\var{p}}+\\var{one}}{\\var{p+1}} \\text{.} \\]
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\ni)
\n\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}}=$ [[0]]
\n\nii)
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}}=$ [[1]]
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\n\n$\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}=$ | \n$\\surd$[[0]] | \n
[[1]] | \n
Schreiben Sie den Nenner als ganze Zahl und kürzen Sie so weit wie möglich.
\n\\[ \\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}} = \\frac{a\\sqrt{\\var{f}}}{b}\\text{,} \\]
\nwobei
\n$a =$ [[0]]
\n$b =$ [[1]]
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\n\n$\\displaystyle\\frac{\\var{g}}{(\\sqrt{\\var{h}}+\\var{j})}$ =
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", "$\\displaystyle-\\var{l}-\\var{o}\\sqrt{\\var{h}}$
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\n\\[ \\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}} = \\frac{a-b\\sqrt{\\var{u}}}{c} \\text{,} \\]
\nwobei:
\n$a = $ [[0]]
\n$b = $ [[1]]
\n$c = $ [[2]]
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\n\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}} \\]
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", "$\\displaystyle\\frac{\\var{p}+\\var{qwerty}\\sqrt{\\var{p}}}{\\var{p}\\sqrt{\\var{p}}}$
", "$\\displaystyle\\frac{\\sqrt{\\var{p}}+\\var{r_coprime*p_p1}}{\\var{p}}$
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