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\(a\left(a+1\right)\left(a+2\right)\left(a+3\right)+1\)
\(=\left(a^2+3a\right)\left(a^2+3a+2\right)+1\)
Đặt \(t=a^2+3a\) thì ta được:
\(t\left(t+2\right)+1=t^2+2t+1\)
\(=\left(t+1\right)^2=\left(a^2+3a+1\right)^2\)
\(VT=a^2+2\left(a^2+2a+1\right)+3\left(a^2+4a+4\right)+4\left(a^2+6a+9\right)\)
\(=a^2+2a^2+4a+2+3a^2+12a+12+4a^2+24a+36\)
\(=10a^2+40a+50=\left(9a^2+30a+25\right)+\left(a^2+10a+25\right)\)
\(=\left(3a+5\right)^2+\left(a+5\right)^2\)
Em kiểm tra lại đề bài nhé!
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
\(\left(a^2-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)=\left(a-1\right)\left(a+1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)=\left[\left(a+1\right)\left(a^2-a+1\right)\right]\left[\left(a-1\right)\left(a^2+a+1\right)\right]=a^3+1+a^3-1=2a^3\)
\(2,\left(x^6-3x^2+9\right)\left(x^3+3\right)=\left(x^3+3\right)\left[\left(x^3\right)^2-3x^2+3^2\right]=x^9+3^3\)