Câu hỏi của Vũ Đình Đức

Question

Cho A=4a(a+b)(a+1)(a+b+1)+`b^2 `
CMR A là số chính phương

Nguyễn Việt Lâm · Accepted Answer

$A=4\left(a^2+a\right)\left[\left(a+b\right)^2+a+b\right]+b^2$$=4a^2\left(a+b\right)^2+4a^2\left(a+b\right)+4a\left(a+b\right)^2+4a\left(a+b\right)+b^2$$=4a^2\left(a+b\right)^2+4a\left(a+b\right)\left(a+b+1\right)+4a^2\left(a+b\right)+b^2$$=4a^2\left(a+b\right)^2+4a^2\left(a+b+1\right)+4ab\left(a+b+1\right)+4a^2\left(a+b\right)+b^2$$=4a^2\left[\left(a+b\right)^2+a+b+1+a+b\right]+4ab\left(a+b+1\right)+b^2$$=4a^2\left[\left(a+b\right)^2+2\left(a+b\right)+1\right]+4ab\left(a+b+1\right)+b^2$$=4a^2\left(a+b+1\right)^2+4ab\left(a+b+1\right)+b^2$$=\left[2a\left(a+b+1\right)+b\right]^2$Câu trả lời: $A=4\left(a^2+a\right)\left[\left(a+b\right)^2+a+b\right]+b^2$$=4a^2\left(a+b\right)^2+4a^2\left(a+b\right)+4a\left(a+b\right)^2+4a\left(a+b\right)+b^2$$=4a^2\left(a+b\right)^2+4a\left(a+b\right)\left(a+b+1\right)+4a^2\left(a+b\right)+b^2$$=4a^2\left(a+b\right)^2+4a^2\left(a+b+1\right)+4ab\left(a+b+1\right)+4a^2\left(a+b\right)+b^2$$=4a^2\left[\left(a+b\right)^2+a+b+1+a+b\right]+4ab\left(a+b+1\right)+b^2$$=4a^2\left[\left(a+b\right)^2+2\left(a+b\right)+1\right]+4ab\left(a+b+1\right)+b^2$$=4a^2\left(a+b+1\right)^2+4ab\left(a+b+1\right)+b^2$$=\left[2a\left(a+b+1\right)+b\right]^2$

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