Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\left(a^8+2a^4b^4+b^8\right)-a^4b^4\)
\(P=\left(a^4+b^4\right)^2-a^4b^4\)
\(P=\left(a^4+b^4+a^2b^2\right)\left(a^4+b^4-a^2b^2\right)\)
\(P=\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^4+b^4-a^2b^2\right)\)
\(P=\left(a^2+b^2+ab\right)\left(a^2+b^2-ab\right)\left(a^4+b^4-a^2b^2\right)\)
\(P=\left[\left(a+b\right)^2-ab\right]\left(a^2+b^2-ab\right)\left(a^4+b^4-a^2b^2\right)\)
\(P=\left(a+b+\sqrt{ab}\right)\left(a+b-\sqrt{ab}\right)\left(a^2+b^2-ab\right)\left(a^4+b^4-a^2b^2\right)\)
Đề sai nhé
\(ab-a-b+1\)
\(=\left(ab-b\right)-\left(a-1\right)\)
\(=b\left(a-1\right)-\left(a-1\right)\)
\(=\left(b+1\right)\left(a-1\right)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)
A= a2(1-b2) - b(1-b) - a(1-b)
=a2(1-b)(1+b) - b(1-b) - a(1-b)
=(1-b)(a2+a2b-b-a)
=(1-b)[a(a-1) + b(a2-1)]= (1-b)[a(a-1) + b(a-1)(a+1) =(1-b)(a-1)(a+ab+b)
Chỗ nào ko hiểu thì hỏi nhé ^^
a^3+b^3+3a^2b+3ab^2+a^3-b^3-3a^2b+3ab^2
=2a^3+6ab^2=2a(a^2+3b^2)
a3+b3+c3-3abc=(a+b)3+c3-3a2b-3ab2-3abc
=(a+b+c)[(a+b)2-(a+b).c+c2]-3ab.(a+b+c)
=(a+b+c)(a2+b2+c2-ac-bc-ab)
\(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)\)
\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)\)
a3+b3+c3−3abca^3+b^3+c^3-3abca3+b3+c3−3abc
=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc
=(a+b)3+c3−(3a2b+3ab2+3abc)=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)=(a+b)3+c3−(3a2b+3ab2+3abc)
=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)
=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)
=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)
=(a+b+c)(a2+b2+c2−ab−ac−ab)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)=(a+b+c)(a2+b2+c2−ab−ac−ab)
a) \(\Rightarrow\left(x^2\right)^2+\left(2^2\right)^2+2.2x^2-2.2x^2\Rightarrow\left(x^2+2\right)^2-\left(2x\right)^2\Rightarrow\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
b) \(\Rightarrow\left(2x^4\right)^2+2.2.x^4.1+1-2.2.x^4.1\Rightarrow\left(2x^4+1\right)^2-\left(2x^2\right)^2\Rightarrow\left(2x^4+1-2x^2\right)\left(2x^4+1+2x^2\right)\)
CHÚC BẠN học tốt
T I C K cho mình nha cảm ơn
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)