Đặt \(M=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(M=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)

\(M=\left(a^2+7a+a+7\right)\left(a^2+5a+3a+15\right)+15\)

\(M=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt \(p=a^2+8a+11\)

\(\Rightarrow M=\left(p-4\right)\left(p+4\right)+15\)

\(\Rightarrow M=p^2-16+15\)

\(\Rightarrow M=p^2-1\)

\(\Rightarrow M=\left(p-1\right)\left(p+1\right)\)

Thay \(p=a^2+8a+11\)vào M, ta có :

\(M=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

\(M=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

Linh Linh con  

A=(x-2)(x^2+2x+4)-x(x^2-2)

=x^3-8-x^3+2x

=2x-8

=x²+2x-3x-6-x²-x-5

=-2x-11

Đúng nha !!!

11) \(a^3+b^3+c^3-3abc\)

\(=a^3+b^3+c^3+3a^2b+3ab^2-3a^2b-3ab^2-3abc\)

\(=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

12) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[a+\left(b+c\right)\right]^3-a^3-b^3-c^3\)

\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+b^3+3b^2c+3bc^2+c^3-a^3-b^3-c^3\)

\(=3a^2\left(b+c\right)+3a\left(b+c\right)+3b^2c+3bc^2\)

\(=3a^2\left(b+c\right)+3a\left(b+c\right)+3bc\left(b+c\right)\)

\(=3\left(b+c\right)\left[a^2+a\left(b+c\right)+bc\right]\)

\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)

\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)

\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)

13) \(2xy-x^2-y^2+16\)

\(=16-\left(x^2-2xy+y^2\right)\)

\(=4^2-\left(x-y\right)^2\)

\(=\left(-x+y+4\right)\left(x-y+4\right)\)

thanks

Danh sách nguyên tố hóa học[sửa | sửa mã nguồn]

Trả lời :

 Nếu tính đến tháng 12 năm 2017

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