// 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. 

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Translated to German, made some minor changes to Advice section.

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Original: 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": "

a)

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Es gilt $\\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\frac{a}{b}}$.

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i)

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$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{{\\var{prime_nums[0]}}}} = \\sqrt{\\var{square_nums[0]}}= \\simplify{{sqrt(square_nums[0])}}$

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ii)

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$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}} = \\sqrt{\\frac{{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{{\\var{prime_nums[1]}}}} = \\sqrt{\\var{square_nums[1]}} = \\simplify{{sqrt(square_nums[1])}}$.

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b)

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Um eine ganze Zahl in den Nenner zu bekommen, benutzen wir, dass $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$.

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Gefragt ist, den Ausdruck $\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}$ umzuschreiben als Bruch mit einer ganzen Zahl im Nenner.

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Der Nenner ist ${\\sqrt{\\var{m_lcm}}}$, also multiplizieren wir den Nenner mit ${\\sqrt{\\var{m_lcm}}}$.

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\\[ \\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}\\cdot\\frac{\\sqrt{\\var{m_lcm}}}{\\sqrt{\\var{m_lcm}}} \\]

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Weil $\\sqrt{a}\\cdot\\sqrt{b}$ = $\\sqrt{ab}$, können wir den Zähler  als $\\sqrt{\\var{num}}$ schreiben.

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Der Nenner ist ${\\sqrt{\\var{m_lcm}}}\\cdot{\\sqrt{\\var{m_lcm}}} = {\\var{m_lcm}}$.

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Also ist das Ergebnis dieser Rechnung

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\\[ \\frac{\\sqrt{\\var{num}}}{\\var{m_lcm}} \\]

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Zum Schluss können wir noch kürzen und erhalten

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\\[ \\frac{\\sqrt{\\var{a[0]*b[0]}}}{\\var{b[0]}} \\]

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(Man kann natürlich auch schon zu Beginn kürzen, das ist im Grunde effizienter.)

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c)

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Zunächst erweitern wir den Bruch mit $\\sqrt{\\var{f}}$. Weil $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$ gilt, ist der Nenner des entstehenden Bruchs $\\var{d}\\cdot\\var{f}=\\var{df}$.

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$\\displaystyle\\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}}\\cdot\\simplify[all,!simplifyFractions,!sqrtDivision]{sqrt({f})/(sqrt({f}))}={\\frac{\\simplify{{c}*sqrt({f})}}{{(\\var{d}\\cdot\\var{f})}}}=\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}$.

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Wir kürzen noch mit $\\var{gcd_cdf}$, und erhalten als Ergebnis

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\\[\\frac{\\simplify{{c}*sqrt({f})}}{\\var{d*f}}=\\frac{\\simplify{{c_coprime}*sqrt({f})}}{\\var{df_coprime}}\\]

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Die Antwort soll in der Form $\\frac{a \\sqrt{\\var{f}}}{b}$ gegeben werden, also ist $a = \\var{c_coprime}$ und $b = \\var{df_coprime}$.

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d)

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Um 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.

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\\[ \\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{.} \\]

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Wir erhalten

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\\[ \\frac{ \\simplify{ {g*-j} + {g}*sqrt({h})} }{\\var{h-j^2}}\\text{,} \\]

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Das können wir noch vereinfachen zu

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\\[ \\var{o}\\sqrt{\\var{h}}-{\\var{l}}\\text{.} \\]

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e)

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Wir rechnen

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$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}$ =

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$\\displaystyle\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}}\\times\\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}-\\sqrt{\\var{u}}}=\\frac{(\\var{t}-\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}{(\\var{t}+\\sqrt{\\var{u}})(\\var{t}-\\sqrt{\\var{u}})}=\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}$.

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$\\displaystyle\\frac{\\var{t^2}-\\var{2t}\\sqrt{\\var{u}}+\\var{u}}{\\var{t^2}-\\var{u}}=\\frac{\\var{t^2+u}-\\var{2t}{\\sqrt{\\var{u}}}}{\\var{t^2-u}}$.

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Kürzen mit ${\\var{gcd_frac}}$ ergibt dann

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$\\displaystyle\\frac{\\var{num_simp_1}-\\simplify{{num_simp_2}*sqrt({u})}}{\\var{denom_simp}}$. 

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f)

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Um 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.

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Beachten 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

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\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\var{p}\\sqrt{\\var{p}}} \\text{.} \\]

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Das vereinfachen wir weiter zu

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\\[ \\frac{\\var{p}}{\\var{p+1}\\sqrt{\\var{p}}} \\text{.} \\]

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Deshalb haben wir als Zwischenergebnis $\\displaystyle\\frac{\\var{p}}{{\\var{p+1}}\\sqrt{\\var{p}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}}$.

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Um die Addition durchzuführen, müssen wir den zweiten Bruch auf denselben Nenner bringen. Wir erweitern mit $\\simplify{{p_p1}sqrt({p})}$, also

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$\\displaystyle\\frac{\\var{r_coprime}}{\\var{s_coprime}}\\cdot\\simplify[all,!simplifyFractions,!sqrtDivision]{{p_p1}*sqrt({p})/({p_p1}*sqrt({p}))}=\\frac{\\simplify{{one}*sqrt({p})}}{\\var{two}\\sqrt{\\var{p}}}$,

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Jetzt erhalten wir

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\\[ \\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}}} \\]

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Weil $\\displaystyle\\sqrt{a}\\cdot\\sqrt{a}=a$ ist, können wir $\\sqrt{\\var{p}}$ im Zähler ausklammern:

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\\[ \\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}\\text{.} \\]

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Wir kürzen jetzt mit $\\sqrt{\\var{p}}$ und bekommen

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\\[ \\frac{\\sqrt{\\var{p}}+\\var{one}}{\\var{p+1}} \\text{.} \\]

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Random number.

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Random numbers, not square.

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Random number.

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Simplified numerical term of numerator.

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GCD of terms in numerator.

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Random numbers, not square.

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List of square numbers.

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Simplified denominator.

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Multiple of LCM of n amd m.

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List of prime numbers.

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Multiple of LCM of n amd m.

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GCD of all terms in fraction.

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Simplified surd term of numerator.

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Vereinfachen Sie die folgenden Ausdrücke und schreiben Sie sie als ganze Zahlen.

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i)

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$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[0]}{prime_nums[0]}}}}{\\sqrt{\\var{prime_nums[0]}}}=$ [[0]]

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ii)

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$\\displaystyle\\frac{\\sqrt{\\simplify{{square_nums[1]}{prime_nums[1]}}}}{\\sqrt{\\var{prime_nums[1]}}}=$ [[1]]

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Vereinfachen Sie den Ausdruck, indem Sie ihn mit einer ganzen Zahl als Nenner schreiben.

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\n\n\n\n\n\n\n\n\n\n\n
$\\displaystyle\\frac{\\sqrt{\\var{n_lcm}}}{\\sqrt{\\var{m_lcm}}}=$$\\surd$[[0]]
[[1]]
\n

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Schreiben Sie den Nenner als ganze Zahl und kürzen Sie so weit wie möglich.

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\\[ \\frac{\\var{c}}{\\var{d}\\sqrt{\\var{f}}} = \\frac{a\\sqrt{\\var{f}}}{b}\\text{,} \\]

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wobei

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$a =$ [[0]]

\n

$b =$ [[1]]

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Schreiben Sie den Ausdruck in der Form $a\\sqrt{b}+m$, mit ganzen Zahlen $a$, $b$ und $c$.

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\n

$\\displaystyle\\frac{\\var{g}}{(\\sqrt{\\var{h}}+\\var{j})}$ = 

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$\\displaystyle\\var{o}\\sqrt{\\var{h}}+\\var{l}$

", "

$\\displaystyle-\\var{l}-\\var{o}\\sqrt{\\var{h}}$

", "

$\\displaystyle\\var{o}\\sqrt{\\var{h}}-\\var{l}$

", "

$\\displaystyle\\var{l}-\\sqrt{\\var{h}}$

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Schreiben Sie mit einer ganzen Zahl als Nenner und vereinfachen Sie so weit wie möglich:

\n

\\[ \\frac{\\var{t}-\\sqrt{\\var{u}}}{\\var{t}+\\sqrt{\\var{u}}} = \\frac{a-b\\sqrt{\\var{u}}}{c} \\text{,} \\]

\n

wobei:

\n

$a = $ [[0]]

\n

$b = $ [[1]]

\n

$c = $ [[2]]

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Vervollständigen Sie diese Addition, indem Sie zuerst den ersten Summanden als Bruch mit einer ganzen Zahl im Nenner ausdrücken. Die Antwort soll ein Bruch mit einer ganzen Zahl im Nenner sein.

\n

\\[ \\frac{\\var{p}}{\\sqrt{\\var{p}}+\\sqrt{\\var{p^3}}}+\\frac{\\var{r_coprime}}{\\var{s_coprime}} \\]

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$\\displaystyle\\frac{\\sqrt{\\var{p}}+\\var{r_coprime*p_p1}}{\\var{p+1}}$

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$\\displaystyle\\frac{\\sqrt{\\var{p}}(\\sqrt{\\var{p}}+\\var{r_coprime*p_p1})}{\\var{p+1}\\sqrt{\\var{p}}}$

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$\\displaystyle\\frac{\\var{p}+\\var{qwerty}\\sqrt{\\var{p}}}{\\var{p}\\sqrt{\\var{p}}}$

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$\\displaystyle\\frac{\\sqrt{\\var{p}}+\\var{r_coprime*p_p1}}{\\var{p}}$

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