Rút gọn biểu thức sau: (a4 - 3a2 +9).(a2 + 3)-(3 +a2)3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có: \(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)
\(=a^6-3a^4+3a^2-1-\left(a^6-1\right)\)
\(=-3a^4+3a^2\)
b: Ta có: \(\left(a^4-3a^2+9\right)\left(a^2+3\right)-\left(a^2+3\right)^3\)
\(=a^6+27-a^6-9a^4-27a^2-27\)
\(=-9a^4-27a^2\)
a,hđt số 3 = \(\left(a^2+2a\right)^2-9\)
b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)
a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
b) \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)
\(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
=1
\(M=\left(a^2+b^2+2-a^2-b^2+2\right)\left[\left(a^2+b^2+2\right)^2+\left(a^2+b^2+2\right)\left(a^2+b^2-2\right)+\left(a^2+b^2-2\right)^2\right]-12\left(a^2+b^2\right)^2\\ M=4\left(a^4+b^4+4+4a^2+4b^2+2a^2b^2+\left(a^2+b^2\right)^2-4+a^4+b^4+4-4a^2-4b^2+2a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2-3a^4-6a^2b^2-3b^4\right)\\ M=4\cdot4=164\)
\(2\sqrt{a^2}=2\left|a\right|=2a\) (với \(a\ge0\) )
\(3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(2-a\right)=6-3a\) (\(a< 2\))
a: \(2\sqrt{a^2}=-2a\)
b: \(3\sqrt{\left(a-2\right)^2}=3\left|a-2\right|=3\left(2-a\right)\)
a) Ta có:
b) Ta có:
c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:
(Vì a ≥ 0 nên |a| = a)
d) Ta có:
\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)
(a4 - 3a2 + 9).(a2 + 3) - (3 + a2)3
= a6 + 27 - 27 - 9a4 - 27a2 - a6
= -9a4 - 27a2
\(\left(a^4-3a^2+9\right)\left(a^2+3\right)-\left(3+a^2\right)^3\)
\(=\left[\left(a^2\right)^3+3^3\right]-\left(27+9a^4+27a^2+a^6\right)\)
\(=a^6+27-27-9a^4-27a^2-a^6\)
\(=-9a^4-27a^2\)