Bài 1: Tìm \(\overline{abcde}\), biết
1) \(\sqrt{\overline{abcde}}\) = 5e + 1
2) \(\sqrt{\overline{abcde}}\) = \(\left(ab\right)^3\)
Bài 2: Cho a, b>0: \(a^{2012}\)+ \(b^{2012}\) = \(a^{2013}\)+\(b^{2013}\)=\(a^{2014}\)+\(b^{2014}\)
Bài 3: Tìm a, b, c: a.( a + b + c ) = \(-\dfrac{1}{24}\)
c.( a + b + c ) = \(-\dfrac{1}{72}\)
b.( a + b + c ) = \(\dfrac{1}{16}\)
(cứu mih với ạ uhuhuhu)
Bài 3.
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=-\dfrac{1}{24}\left(1\right)\\c\left(a+b+c\right)=-\dfrac{1}{72}\left(2\right)\\b\left(a+b+c\right)=\dfrac{1}{16}\left(3\right)\end{matrix}\right.\)
Dễ thấy \(a,b,c\ne0\Rightarrow a+b+c\ne0\)
Chia (1) cho (2), ta được \(\dfrac{a}{c}=3\Rightarrow a=3c\left(4\right)\)
Chia (2) cho (3) ta được: \(\dfrac{c}{b}=-\dfrac{2}{9}\Rightarrow b=-\dfrac{9}{2}c\left(5\right)\).
Thay (4), (5) vào (2), ta được: \(-\dfrac{1}{2}c^2=-\dfrac{1}{72}\)
\(\Rightarrow c=\pm\dfrac{1}{6}\).
Với \(c=\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=\dfrac{1}{2}\\b=-\dfrac{9}{2}c=-\dfrac{3}{4}\end{matrix}\right.\)
Với \(c=-\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=-\dfrac{1}{2}\\b=-\dfrac{9}{2}c=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(\left(a;b;c\right)=\left\{\left(\dfrac{1}{2};-\dfrac{3}{4};\dfrac{1}{6}\right);\left(-\dfrac{1}{2};\dfrac{3}{4};-\dfrac{1}{6}\right)\right\}\)
