Question

Va rog , exercitiul 19 punctul b)

Answer 1

Daca $\sqrt{ n^{2} -7n+5}$∈N => $2 \sqrt{ n^{2}-7n+5 } = \sqrt{4 n^{2}-28n+20 }$$= \sqrt{(4 n^{2}-28n+49)-29}= \sqrt{(2n-7) ^{2} -29}$∈N.

Fie k∈Z a.i. $\sqrt{(2n-7) ^{2}-29 } = k=\ \textgreater \ (2n-7)^{2}-29= k^{2} =\ \textgreater \ \\ \\ =\ \textgreater \ (2n-7) ^{2} - k^{2} =29=1*29=(-1)*(-29).$

Rezolvam urmatoarele sisteme de ecuatii prin metoda reducerii :

$\left \{ {{2n-7+k=1} \atop {2n-7-k=29}} \right. ; \left \{ {{2n-7+k=-1} \atop {2n-7-k=-29}} \right. ; \left \{ {{2n-7+k=29} \atop {2n-7-k=1}} \right.~si~ \left \{ {{2n-7+k=-29} \atop {2n-7-k=-1}} \right. .$

Primul si ultimul sistem ne conduc la 2k=-28=>k=-14. Al doilea si al treilea sistem ne conduc 2k=28=>k=14.

Deci k=14 sau k=-14.

$k^{2} =196.$

Avem, asadar: $(2n-7) ^{2} -29= k^{2} \ \textless \ =\ \textgreater \ (2n-7) ^{2} -29=196=\ \textgreater \ \\ =\ \textgreater \ (2n-7)=225=\ \textgreater \ 2n-7=15~SAU~2n-7=-15. \\ \\ 2n-7=15=\ \textgreater \ 2n=22=\ \textgreater \ \boxed{n=11}. \\ 2n-7=-15=\ \textgreater \ 2n=-8=\ \textgreater \ \boxed{n=-4}.$