Ummm x fa ayudenme con este problema ...ahh si no me aduyan x fa les suplico que no den respuestas fuera del tema ...muchas gracias ...26 puntos
yexs
![\frac{a(a+1)}{2}=b=\ \textgreater \ \frac{b(b+1)}{2}~,-\ \textgreater \ nuevamente~este~ultimo~igualamos~a ''c'' \\ \frac{b(b+1)}{2}=c~~--\ \textgreater \ Usemos~la~misma~condicion~que~es: \\ \frac{c(c+1)}{2} , Como~ya~hicimos~3~operaciones, quiere~decir~hay~3~triangulos \\ en~la~figura~este~ultimo~igualamos~a~21~veamos: \\ \\ \frac{c(c+1)}{2}=21 \\ c(c+1)=42~~--\ \textgreater \ busquemos~2~numeros~que~multiplique \\ c(c+1)=6_X7~~--\ \textgreater \ entonces~[c=6] \\ \boxed{c=6} , como~ya~tenemos~valor~de~''c'', calculemos~la~''b'' \\ \\ \frac{b(b+1)}{2}=c~~ \\ \\](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%28a%2B1%29%7D%7B2%7D%3Db%3D%5C+%5Ctextgreater+%5C++%5Cfrac%7Bb%28b%2B1%29%7D%7B2%7D%7E%2C-%5C+%5Ctextgreater+%5C+nuevamente%7Eeste%7Eultimo%7Eigualamos%7Ea+%27%27c%27%27+%5C%5C++%5Cfrac%7Bb%28b%2B1%29%7D%7B2%7D%3Dc%7E%7E--%5C+%5Ctextgreater+%5C+Usemos%7Ela%7Emisma%7Econdicion%7Eque%7Ees%3A+%5C%5C+++%5Cfrac%7Bc%28c%2B1%29%7D%7B2%7D++%2C+Como%7Eya%7Ehicimos%7E3%7Eoperaciones%2C+quiere%7Edecir%7Ehay%7E3%7Etriangulos+%5C%5C+en%7Ela%7Efigura%7Eeste%7Eultimo%7Eigualamos%7Ea%7E21%7Eveamos%3A+%5C%5C++%5C%5C++%5Cfrac%7Bc%28c%2B1%29%7D%7B2%7D%3D21+%5C%5C+c%28c%2B1%29%3D42%7E%7E--%5C+%5Ctextgreater+%5C+busquemos%7E2%7Enumeros%7Eque%7Emultiplique+%5C%5C+c%28c%2B1%29%3D6_X7%7E%7E--%5C+%5Ctextgreater+%5C+entonces%7E%5Bc%3D6%5D+%5C%5C+%5Cboxed%7Bc%3D6%7D+%2C+como%7Eya%7Etenemos%7Evalor%7Ede%7E%27%27c%27%27%2C+calculemos%7Ela%7E%27%27b%27%27+++%5C%5C++%5C%5C++%5Cfrac%7Bb%28b%2B1%29%7D%7B2%7D%3Dc%7E%7E+++%5C%5C++%5C%5C++ "\frac{a(a+1)}{2}=b=\ \textgreater \ \frac{b(b+1)}{2}~,-\ \textgreater \ nuevamente~este~ultimo~igualamos~a ''c'' \\ \frac{b(b+1)}{2}=c~~--\ \textgreater \ Usemos~la~misma~condicion~que~es: \\ \frac{c(c+1)}{2} , Como~ya~hicimos~3~operaciones, quiere~decir~hay~3~triangulos \\ en~la~figura~este~ultimo~igualamos~a~21~veamos: \\ \\ \frac{c(c+1)}{2}=21 \\ c(c+1)=42~~--\ \textgreater \ busquemos~2~numeros~que~multiplique \\ c(c+1)=6_X7~~--\ \textgreater \ entonces~[c=6] \\ \boxed{c=6} , como~ya~tenemos~valor~de~''c'', calculemos~la~''b'' \\ \\ \frac{b(b+1)}{2}=c~~ \\ \\")
![\frac{b(b+1)}{2}=6 \\ b(b+1)=12~~--\ \textgreater \ sabemos~que~3_X4=12 \\ b(b+1)=3_X4~~--\ \textgreater \ igualando \\ \boxed{b=3}} \\ Ahora~calculemos~valor~de~''a'', veamos: \\ \frac{a(a+1)}{2}=b~~~--\ \textgreater \ se~sabe~que~[b=3], reemplazando \\ \frac{a(a+1)}{2}=3 \\ a(a+1)=6~~--\ \textgreater \ 6~es~igual~a~2_X3 \\ \boxed{a=2} \\ \\ Y~por~ultimo~vamos~calcular~''x''~veamos: \\ \\ 2x+1=a~~--\ \textgreater \ [a=2], reemplazando \\ 2x+1=2 \\ 2x=2-1 \\ 2x=1 \\ \\ \boxed{\boxed{\boxed{x= \frac{1}{2}}}}~~--\ \textgreater \ respuesta~alternativa~\textcircled{b} \\ \\ \\ .](https://tex.z-dn.net/?f=++%5Cfrac%7Bb%28b%2B1%29%7D%7B2%7D%3D6+%5C%5C+b%28b%2B1%29%3D12%7E%7E--%5C+%5Ctextgreater+%5C+sabemos%7Eque%7E3_X4%3D12+%5C%5C+b%28b%2B1%29%3D3_X4%7E%7E--%5C+%5Ctextgreater+%5C+igualando+%5C%5C+%5Cboxed%7Bb%3D3%7D%7D++%5C%5C+Ahora%7Ecalculemos%7Evalor%7Ede%7E%27%27a%27%27%2C+veamos%3A+%5C%5C++%5Cfrac%7Ba%28a%2B1%29%7D%7B2%7D%3Db%7E%7E%7E--%5C+%5Ctextgreater+%5C+se%7Esabe%7Eque%7E%5Bb%3D3%5D%2C+reemplazando+%5C%5C++%5Cfrac%7Ba%28a%2B1%29%7D%7B2%7D%3D3+%5C%5C+a%28a%2B1%29%3D6%7E%7E--%5C+%5Ctextgreater+%5C+6%7Ees%7Eigual%7Ea%7E2_X3++%5C%5C+%5Cboxed%7Ba%3D2%7D++%5C%5C++%5C%5C+Y%7Epor%7Eultimo%7Evamos%7Ecalcular%7E%27%27x%27%27%7Eveamos%3A+%5C%5C++%5C%5C+2x%2B1%3Da%7E%7E--%5C+%5Ctextgreater+%5C+%5Ba%3D2%5D%2C+reemplazando+%5C%5C+2x%2B1%3D2+%5C%5C+2x%3D2-1+%5C%5C+2x%3D1+%5C%5C++%5C%5C+%5Cboxed%7B%5Cboxed%7B%5Cboxed%7Bx%3D+%5Cfrac%7B1%7D%7B2%7D%7D%7D%7D%7E%7E--%5C+%5Ctextgreater+%5C+respuesta%7Ealternativa%7E%5Ctextcircled%7Bb%7D+%5C%5C++%5C%5C++%5C%5C+. "\frac{b(b+1)}{2}=6 \\ b(b+1)=12~~--\ \textgreater \ sabemos~que~3_X4=12 \\ b(b+1)=3_X4~~--\ \textgreater \ igualando \\ \boxed{b=3}} \\ Ahora~calculemos~valor~de~''a'', veamos: \\ \frac{a(a+1)}{2}=b~~~--\ \textgreater \ se~sabe~que~[b=3], reemplazando \\ \frac{a(a+1)}{2}=3 \\ a(a+1)=6~~--\ \textgreater \ 6~es~igual~a~2_X3 \\ \boxed{a=2} \\ \\ Y~por~ultimo~vamos~calcular~''x''~veamos: \\ \\ 2x+1=a~~--\ \textgreater \ [a=2], reemplazando \\ 2x+1=2 \\ 2x=2-1 \\ 2x=1 \\ \\ \boxed{\boxed{\boxed{x= \frac{1}{2}}}}~~--\ \textgreater \ respuesta~alternativa~\textcircled{b} \\ \\ \\ .")
1 votes
Thanks 1
newgirls
gracias yex