Problemă rezolvată dintre subiectele propuse pentru clasa a IX-a la ONGM 2020-2021.
Enunț
Să se afle valoarea expresiei E(x) :
\displaystyle E(x)= \dfrac{\sqrt[]{x+2}-\sqrt[]{x-2}}{\sqrt[]{x+2}+\sqrt[]{x-2}}știind că:
{a\in (0,1)} \;\;\;\;și\;\;\;\; \displaystyle x = a + \dfrac{1}{a}Rezolvare:
În prima etapă, amplificăm fracția cu
\sqrt[]{x+2}-\sqrt[]{x-2}și obținem:
E(x)= \frac{\sqrt[]{x+2}-\sqrt[]{x-2}}{\sqrt[]{x+2}+\sqrt[]{x-2}} \Leftrightarrow \\\;\\
\Leftrightarrow E(x)= \dfrac{(\sqrt[]{x+2}-\sqrt[]{x-2})^2}{(\sqrt[]{x+2}+\sqrt[]{x-2})\cdot(\sqrt[]{x+2}-\sqrt[]{x-2})} \Leftrightarrow \\\;\\
\Leftrightarrow E(x)= \dfrac{(\sqrt[]{x+2}-\sqrt[]{x-2})^2}{(\sqrt[]{x+2})^2-(\sqrt[]{x-2})^2} \Leftrightarrow \\\;\\
\Leftrightarrow
E(x)= \dfrac{(\sqrt[]{x+2}-\sqrt[]{x-2})^2}{{({x}+2)}-{({x}-2)}} \Leftrightarrow\\\;\\
\Leftrightarrow \displaystyle E(x)= \dfrac{(\sqrt[]{x+2})^2-2\cdot\sqrt[]{x+2}\cdot\sqrt[]{x-2}+(\sqrt[]{x-2})^2}{{\cancel x+2}-{\cancel x+2}} \Leftrightarrow \\\;\\
\Leftrightarrow \displaystyle E(x)= \dfrac{{x+\cancel{2}}-2\cdot\sqrt[]{(x+2)\cdot(x-2)}+{x-\cancel{2}}}{{2+2}} \Leftrightarrow \\\;\\
\Leftrightarrow \displaystyle E(x)= \dfrac{x-2\cdot\sqrt[]{x^2-2^2}+{x}}{{4}}
\\\;\\ \Leftrightarrow
\displaystyle E(x)= \dfrac{2\cdot x-2\cdot\sqrt[]{x^2-4}}{{4}} \Leftrightarrow
\\\;\\
\Leftrightarrow\displaystyle E(x)= \dfrac{2\cdot( x-\sqrt[]{x^2-4})}{{4}} \Leftrightarrow \\\;\\\Leftrightarrow
\displaystyle E(x)= \dfrac{x-\sqrt[]{x^2-4}}{{2}}
[the_ad_group id=”103″]
Știm că:
\displaystyle x = a + \dfrac{1}{a} \implies \displaystyle x^2 = \left(a + \dfrac{1}{a}\right)^2 \implies \\\;\\\ \implies \displaystyle x^2 = a^2 + 2\cdot a\cdot \dfrac{1}{a}+ \left(\dfrac{1}{a}\right)^2 \implies \\\;\\\\
\implies\displaystyle x^2 = a^2 + 2\cdot \cancel a\cdot \dfrac{1}{\cancel a}+ \left(\dfrac{1}{a}\right)^2 \implies\\\;\\\
\implies\displaystyle x^2 = a^2 + 2 + \left(\dfrac{1}{a}\right)^2 \implies \\\;\\\
\implies \displaystyle x^2 -4 = a^2 + 2 + \left(\dfrac{1}{a}\right)^2 -4 \implies\\\;\\\
\implies
\displaystyle x^2 -4 = a^2 - 2 + \left(\dfrac{1}{a}\right)^2 \implies\\\;\\\
\implies \displaystyle x^2 -4 = a^2 - 2\cdot a\cdot \dfrac{1}{a}+ \left(\dfrac{1}{a}\right)^2 \implies \\\;\\\
\implies
\displaystyle x^2 -4 = \left(a - \dfrac{1}{a} \right)^2
Înlocuind acum x și x2-4 în E(x), obținem:
\displaystyle E(x)= \dfrac{x-\sqrt[]{x^2-4}}{{2}}\implies\\\;\\\implies\displaystyle E(x)= \dfrac{a + \dfrac{1}{a}-\sqrt[]{\left(a - \dfrac{1}{a} \right)^2}}{{2}} \implies \\\;\\\\ \implies
\displaystyle E(x)= \dfrac{a + \dfrac{1}{a}-\Big|{a - \dfrac{1}{a} }\Big|}{{2}} Cum a este între 0 și 1, explicitarea modulului se face după cum urmează:
a\in (0,1) \implies a<1 \ \overset{a\not= 0}{\implies}\\\;\\
\overset{a\not= 0}{\implies} \left\{
\begin{array}{ll}
1<\dfrac{1}{a}\Big|-a \\\;\\
0<1-a
\end{array}
\right \} \implies \\\;\\ {\implies}
0<1-a<\dfrac{1}{a} -a Rezultă deci:
\displaystyle E(x)= \dfrac{a + \dfrac{1}{a}-\left({ \dfrac{1}{a} -a }\right)}{{2}} {\implies} \\\;\\
\implies \displaystyle E(x)= \dfrac{\cancel 2 \cdot a }{\cancel 2} \implies \\\;\\
\implies\displaystyle \bm{E(x)= a} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[the_ad_group id=”102″]
