Exercices : Racines carrées – 3ac

Exercice 1 Mathxi math

- Sans calculatrice [Signaler une erreur]

Sans calculatrice, donner la valeur exacte des nombres suivants :

\begin{array}{ll}
\mathbf{1.}\ A = \sqrt {1000000}&\quad\mathbf{2.}\ B = \dfrac{{\sqrt 9 + \sqrt {121} }}{{\sqrt {49} }}\\
\mathbf{3.}\ C = \sqrt {\dfrac{{50}}{{98}}}&\quad\mathbf{4.}\ D = \sqrt {31 + \sqrt {21 + \sqrt {9 + \sqrt {49} } } }\\
\end{array}

Indication math

Corrigé

Exercice 2 Mathxi math

- Écrire sous la forme $a\sqrt b$ [Signaler une erreur]

Enoncé

Simplifier et calculer :
\begin{array}{ll}
\mathbf{1.}\ A = \sqrt {50} &\quad\mathbf{2.}\ B = \sqrt {363}\\
\mathbf{3.}\ C = 5\sqrt {27} &\quad\mathbf{4.}\ D = \sqrt {24} + 7\sqrt 6 + 2\sqrt {54}\\
\mathbf{5.}\ E = \sqrt 3 \times \sqrt {21} \times \sqrt 7 &\quad\mathbf{6.}\ F = \sqrt {{5^3} \times {7^5} \times 1000}\\
\mathbf{7.}\ G = \sqrt {242} \times \sqrt {128} &\quad\mathbf{8.}\ H = \sqrt 7 \left[ {\sqrt {700} + {{\left( {\sqrt 7 } \right)}^3}} \right]\\
\mathbf{9.}\ I = \left( {\sqrt {13} – 5} \right)\left( {\sqrt {13} + 5} \right) &\quad\mathbf{10.}\ J = {\left( {\sqrt 5 + 2} \right)^2} \\
\mathbf{11.}\ K = {\left( {\sqrt 3 – 1} \right)^4} &\quad\mathbf{12.}\ L = \left( {\sqrt 3 + 5} \right)\left( {2\sqrt 3 + 1} \right)\\
\mathbf{13.}\ M = \sqrt {{{\left( {\sqrt 7 – 3} \right)}^2}} &\quad\mathbf{14.}\ N = \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 – 7} \right)}^2}}\ \\
\mathbf{15.}\ O = \sqrt {3 + 2\sqrt 2 } &\quad\mathbf{16.}\ P = \sqrt {8 – 2\sqrt {12} }\\
\end{array}

Indication math

Corrigé

$\mathbf{1.}\quad A = \sqrt {50}=\sqrt {25\times 2}=\sqrt {25}\times \sqrt {2}=5\sqrt {2}.$

$\mathbf{2.}\quad B=\sqrt{363}=\sqrt{121\times 3}=\sqrt{121}\times\sqrt{3}=11\sqrt{3}.$

$\mathbf{3.}\quad C = 5\sqrt{27}=5\sqrt{9\times 3}=5\sqrt{9}\times \sqrt{3}=5\times3\sqrt{3}=15\sqrt{3}.$

$\mathbf{4.}\quad D =\sqrt{24}+7\sqrt{6}+2\sqrt{54}=\sqrt{4\times6}+7\sqrt{6}+2\sqrt{9\times6}=2\sqrt{6}+7\sqrt{6}+6\sqrt{6}=15\sqrt{6}.$

$\mathbf{5.}\quad E = \sqrt 3 \times \sqrt {21} \times \sqrt 7 =
\sqrt 3 \times \sqrt {3\times 7} \times \sqrt 7=
\sqrt 3 \times \sqrt {3}\times \sqrt{7} \times \sqrt 7=3\times7=21.$

$\begin{aligned}
\mathbf{6.}\quad F&=\sqrt{{5^3}\times{7^5}\times 1000}=\sqrt{5^2\times(7^2)^2\times10^2\times10\times5}=5\times7^2\times10\sqrt{50}\\
&=2450\sqrt{5^2\times 2}=2450\times5\sqrt{2}=12252\sqrt{2}.\end{aligned}$

$\mathbf{7.}\quad G = \sqrt {242} \times \sqrt {128}=\sqrt{11^2\times2}\times\sqrt{8^2\times 2}=11\sqrt{2}\times8\sqrt{2}=88\sqrt{2}^2=88\times2=176.$

$\begin{aligned}
\mathbf{8.}\quad H &= \sqrt 7 \left[ {\sqrt {700} + {{ {\sqrt 7 } }^3}} \right]=\sqrt 7 \left[ \sqrt {10^2\times7} + { \sqrt{7}^2 \times\sqrt{7} } \right]=
\sqrt 7 \left[ 10\sqrt {7} + { 7 \sqrt{7} } \right]\\
&=10\sqrt{7}^2+7\sqrt{7}^2=10\times 7 + 7\times 7 = 70+49=119.\end{aligned}$

$\mathbf{9.}\quad I = \left( {\sqrt {13} – 5} \right)\left( {\sqrt {13} + 5} \right)=\sqrt{13}^2-5^2=13-25=-12.$

$\mathbf{10.}\quad J = {\left( {\sqrt 5 + 2} \right)^2}=\sqrt{5}^2+2\times\sqrt{5}\times2+2^2=5+4\sqrt{5}+4=9+4\sqrt{5}.$

$\begin{aligned}
\mathbf{11.}\quad K&=\left(\sqrt 3 – 1 \right)^4=\left[\left(\sqrt 3 – 1 \right)^2\right]^2=\left[\sqrt{3}^2-2\sqrt{3}+1^2\right]^2=\left[3-2\sqrt{3}+1\right]^2\\
&=\left[4-2\sqrt{3}\right]^2=4^2-2\times4\times2\sqrt{3}+(2\sqrt{3})^2=16+16\sqrt{3}
+4\times3=28-16\sqrt{3}.
\end{aligned}$

$\mathbf{12.}\quad L = \left( {\sqrt 3 + 5} \right)\left( {2\sqrt 3 + 1} \right)=2\sqrt{3}^2+\sqrt{3}+10\sqrt{3}+5=2\times3+11\sqrt{3}+5=11+11\sqrt{3}.$

$\mathbf{13.}\quad M = \sqrt {{{\left( {\sqrt 7 – 3} \right)}^2}}=-\left( \sqrt 7 – 3 \right)=3-\sqrt{7}$, car $\left( \sqrt 7 – 3 \right)$ est négatif.

Remarque : Si $a$ est négatif, alors $\sqrt{a^2}=-a$.

$\mathbf{14.}\quad$ On a : $\sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}}=\left( {\sqrt 5 – 1} \right)$, car $\left( {\sqrt 5 – 1} \right)$ est positif.

On a : $\sqrt {{{\left( {\sqrt 5 – 7} \right)}^2}}=-\left( {\sqrt 5 – 7} \right)$, car $\left( {\sqrt 5 – 7} \right)$ est négatif.

Donc : $N = \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 – 7} \right)}^2}}=\left( {\sqrt 5 – 1} \right)-\left( {\sqrt 5 – 7} \right)=6$.

$\mathbf{15.}\quad O = \sqrt {3 + 2\sqrt 2 }=\sqrt{\sqrt{2}^2+2\sqrt{2}+1^2}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1$, (car $\sqrt{2}+1>0$).

$\mathbf{16.}\quad P = \sqrt {8 – 2\sqrt {12}}=\sqrt{\sqrt{6}^2-2.\sqrt{6}.\sqrt{2}+\sqrt{2}^2}=\sqrt{(\sqrt{6}-\sqrt{2})^2}=\sqrt{6}-\sqrt{2}$, car $(\sqrt{6}>\sqrt{2}).$

Exercice 3 Mathxi math

- Rationaliser le dénominateur [Signaler une erreur]

Enoncé

Écrire sans « $\sqrt{~~}$ » au dénominateur.

\begin{array}{ll}
\mathbf{1.}\ A = \dfrac{3}{{\sqrt 2 – 1}} &\quad\mathbf{2.}\ B = \dfrac{{\sqrt 5 – 3}}{{\sqrt 5 }}\\
\mathbf{3.}\ C = \dfrac{5}{{\sqrt 7 – 2}} – \dfrac{2}{{\sqrt 7 }} &\quad\mathbf{4.}\ D = \dfrac{{3 + \sqrt 5 }}{{7 + \sqrt 5 }} – \dfrac{{3 – \sqrt 5 }}{{7 – \sqrt 5 }}\\
\mathbf{5.}\ E = \dfrac{{\sqrt 2 + \sqrt 3 }}{{\sqrt 5 – \sqrt 7 }} &\quad\mathbf{6.}\ F = \dfrac{1}{{1 + \sqrt 2 + \sqrt 3 }}\\
\end{array}

Indication math

Corrigé

$\mathbf{1.}\quad A = \dfrac{3}{{\sqrt 2 – 1}} = \dfrac{{3\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 – 1} \right)\left( {\sqrt 2 + 1} \right)}} = \dfrac{{3\sqrt 2 + 3}}{{{{\sqrt 2 }^2} – {1^2}}} = \dfrac{{3\sqrt 2 + 3}}{{2 – 1}} =3\sqrt 2 + 3$

$\mathbf{2.}\quad B = \dfrac{{\sqrt 5 – 3}}{{\sqrt 5 }} = \dfrac{{\left( {\sqrt 5 – 3} \right) \times \sqrt 5 }}{{\sqrt 5 \times \sqrt 5 }} = \dfrac{{\sqrt{5}^2 – 3\sqrt 5 }}{{{{\sqrt 5 }^2}}} =\dfrac{{5 – 3\sqrt 5 }}{5}$

$\begin{aligned}
\mathbf{3.}\quad C &= \dfrac{5}{{\sqrt 7 – 2}} – \dfrac{2}{{\sqrt 7 }} = \dfrac{{5\sqrt 7 – 2\sqrt 7 + 4}}{{\left( {\sqrt 7 – 2} \right)\sqrt 7 }} = \dfrac{{4 + 3\sqrt 7 }}{{7 – 2\sqrt 7 }} = \dfrac{{\left( {4 + 3\sqrt 7 } \right)\left( {7 + 2\sqrt 7 } \right)}}{{\left( {7 – 2\sqrt 7 } \right)\left( {7 + 2\sqrt 7 } \right)}}\\
&= \dfrac{{28 + 8\sqrt 7 + 21\sqrt 7 + 6 \times 7}}{{{7^2} – {{\left( {2\sqrt 7 } \right)}^2}}} =\dfrac{{70 + 29\sqrt 7 }}{{21}}
\end{aligned}$

$\begin{aligned}
\mathbf{4.}\quad D &= \dfrac{{3 + \sqrt 5 }}{{7 + \sqrt 5 }} – \dfrac{{3 – \sqrt 5 }}{{7 – \sqrt 5 }} = \dfrac{{\left( {3 + \sqrt 5 } \right)\left( {7 – \sqrt 5 } \right) – \left( {3 – \sqrt 5 } \right)\left( {7 + \sqrt 5 } \right)}}{{\left( {7 + \sqrt 5 } \right)\left( {7 – \sqrt 5 } \right)}} \\
&= \dfrac{{\left( {21 – 3\sqrt 5 + 7\sqrt 5 – 5} \right) – \left( {21 + 3\sqrt 5 – 7\sqrt 5 – 5} \right)}}{{{7^2} – {{\sqrt 5 }^2}}} = \dfrac{{8\sqrt 7 }}{44} = \dfrac{4\sqrt 7}{11}
\end{aligned}$

$\begin{aligned}
\mathbf{5.}\quad E &= \dfrac{{\sqrt 2 + \sqrt 3 }}{{\sqrt 5 – \sqrt 7 }} = \dfrac{{\left( {\sqrt 2 + \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 7 } \right)}}{{\left( {\sqrt 5 – \sqrt 7 } \right)\left( {\sqrt 5 + \sqrt 7 } \right)}} = \dfrac{{\left( {\sqrt 2 + \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 7 } \right)}}{{{{\sqrt 5 }^2} – {{\sqrt 7 }^2}}}\\
&= \dfrac{{\left( {\sqrt 2 + \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 7 } \right)}}{{5 – 7}} =\dfrac{{\left( {\sqrt 2 + \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 7 } \right)}}{{ – 2}}
\end{aligned}$

$\begin{aligned}
\mathbf{6.}\quad F &= \dfrac{1}{{1 + \sqrt 2 + \sqrt 3 }} = \dfrac{{\left( {1 + \sqrt 2 } \right) – \sqrt 3 }}{{\left[ {\left( {1 + \sqrt 2 } \right) + \sqrt 3 } \right]\left[ {\left( {1 + \sqrt 2 } \right) – \sqrt 3 } \right]}}= \dfrac{{1 + \sqrt 2 – \sqrt 3 }}{{{{\left( {1 + \sqrt 2 } \right)}^2} – {{\sqrt 3 }^2}}} \\
&= \dfrac{{1 + \sqrt 2 – \sqrt 3 }}{{{1^2} + 2\sqrt 2 + {{\sqrt 2 }^2} – {{\sqrt 3 }^2}}}
= \dfrac{{1 + \sqrt 2 – \sqrt 3 }}{{1 + 2\sqrt 2 + 2 – 3}} = \dfrac{{1 + \sqrt 2 – \sqrt 3 }}{{2\sqrt 2 }}\\
&= \dfrac{{\left( {1 + \sqrt 2 – \sqrt 3 } \right) \times \sqrt 2 }}{{2\sqrt 2 \times \sqrt 2 }}
=\dfrac{\sqrt{2}+2-\sqrt{6}}{2\times 2}= \dfrac{{2 +\sqrt 2 – \sqrt 6 }}{4}
\end{aligned}$