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Đặt a + b = x, \(x^3\) là số có 3 chữ số
=> 100 ≤ \(x^3\) ≤ 999 <=> 5 ≤ x ≤ 9
Vậy \(x^3\)phải có dạng aba => x = 7 => a = 3, b = 4 (\(7^3\) = 343)
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a) aba x aa = aaaa
aba = aaaa : aa
aba = 101
=> a = 1 ; b = 0
b) a x b x ba = aaa
b x ba = aaa : a
b x ba = 111
=> a=1
b x b1 = 111
b x b x 10 +1 = 111
b x b x 10 = 111 -1
b x b x 10 = 110
b x b = 110 : 10
b xb = 11
=> b = 3 , 316625
Ta có:
a, \(aba\times aa=aaaa\)
\(\Leftrightarrow aba=aaaa\div aa\)
\(\Leftrightarrow aba=101\)
\(\Rightarrow a=1\); \(b=0\)
Vậy \(a=1\)và \(b=0\)
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ab x aba = abab
aba = ab x 101 : ab
aba = 101
=> \(\hept{\begin{cases}a=1\\b=0\end{cases}}\)
198 đó