Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(a+b+c\right)^3+\left(a-b-c\right)^3-6a\left(b+c\right)^2\)
\(=\left[a+\left(b+c\right)\right]^3+\left[a-\left(b+c\right)\right]^3-6a\left(b+c\right)^2\)
\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3+a^3-3a^2\left(b+c\right)+3a\left(b+c\right)^2-\left(b+c\right)^3-6a\left(b+c\right)^2\)
\(=2a^3\)
\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)
a) (a+b)3+(a-b)3=a3+3a2b+3ab2+b3+a3-3a2b+3ab2-b3
=2a3+6ab2
b) (a + b + c)2 + (a − b − c)2 + (b − c − a)2 + (c − a − b)2
=a2+b2+c2+2ab+2bc+2ca+a2+b2+c2-2ab+2bc-2ac+a2+b2+c2-2bc+2ca-2ba+a2+b2+c2-2ca+2ab-2cb
=4a2+4b2+4c2
a) Ta có: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\cdot\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\cdot\left(a^2+3b^2\right)\)
\(=2a^3+6ab^2\)
\(a,\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(=2b^3\)
\(b,\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)
\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-6ab^2\)
\(=2a^3\)
Câu hỏi của Adolf Hitler - Toán lớp 8 - Học toán với OnlineMath
\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3+a^3-3a^2\left(b+c\right)+3a\left(b+c\right)^2-\left(b+c\right)^3-6a\left(b+c\right)^2\)
\(=2a^3+6a\left(b+c\right)^2-6a\left(b+c\right)^2\)
\(=2a^3\)