a) x = [((n + 1)(n + 4)].[(n + 2)(n + 3)] + 1
= (n2 + 5n + 4)(n2 + 5n + 6) + 1
= (n2 + 5n + 5 - 1)(n2 + 5n + 5 + 1) + 1
= (n2 + 5n + 5)2 - 12 + 1 = (n2 + 5n + 5)2 (đpcm)
b) y = [n(n + 9)].[(n + 3)(n + 6)] + 81
= (n2 + 9n).(n2 + 9n + 18) + 81
= (n2 + 9n + 9 - 9)(n2 + 9n + 9 + 9) + 81
= (n2 + 9n + 9)2 - 92 + 81 = (n2 + 9n + 9)2 (đpcm)
a) \(x=\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)+1\)
\(=\left(n+1\right)\left(n+4\right)\left(n+2\right)\left(n+3\right)+1\)
\(=\left(n^2+5n+4\right)\left(n^2+5n+6\right)+1\) ( 1 )
Đặt \(t=n^2+5n\)
\(\left(1\right)\Leftrightarrow=\left(t+4\right)\left(t+6\right)+1\)
\(=t^2+10+24+1\)
\(=t^2+10t+25\)
\(=\left(t+5\right)^2\)
Vậy x là số chính phương
b) \(y=n\left(n+3\right)\left(n+6\right)\left(n+9\right)+81\)
\(=n\left(n+9\right)\left(n+3\right)\left(n+6\right)+81\)
\(=\left(n^2+9n\right)\left(n^2+9n+18\right)+81\) ( 1 )
Đặt \(a=n^2+9n\)
\(\Leftrightarrow\left(1\right)=a\left(a+18\right)+81\)
\(=a^2+18a+81\)
\(=\left(a+9\right)^2\)
Vậy y là số chính phương
a) Ta có: \(x=\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)+1\)
\(\Leftrightarrow x=\left[\left(n+1\right)\left(n+4\right)\right].\left[\left(n+2\right)\left(n+3\right)\right]+1\)
\(\Leftrightarrow x=\left(n^2+5n+4\right).\left(n^2+5n+6\right)+1\)
Đặt \(a=n^2+5n+4\)\(\Rightarrow\)\(a+2=n^2+5n+6\)
Ta lại có: \(x=a.\left(a+2\right)+1\)
\(\Leftrightarrow x=a^2+2a+1\)
\(\Leftrightarrow x=\left(a+1\right)^2\)
\(\Leftrightarrow x=\left(n^2+5n+5\right)^2\)
Vậy x là số chính phương
b) Ta có: \(y=n\left(n+3\right)\left(n+6\right)\left(n+9\right)+81\)
\(\Leftrightarrow y=\left[n\left(n+9\right)\right]\left[\left(n+3\right)\left(n+6\right)\right]+81\)
\(\Leftrightarrow y=\left(n^2+9n\right)\left(n^2+9n+18\right)+81\)
Đặt \(b=n^2+9n\)\(\Rightarrow\)\(b+18=n^2+9n+18\)
Ta có: \(y=b.\left(b+18\right)+81\)
\(\Leftrightarrow y=b^2+18b+81\)
\(\Leftrightarrow y=\left(b+9\right)^2\)
\(\Leftrightarrow y=\left(n^2+9n+9\right)^2\)
Vậy y là số chính phương
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