Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu hỏi tương tự
Điểm sao thì ít
Ngồi làm thì nhiều
Ai cho sao tôi
Thì thương tôi với
Câu a) dễ, ko làm
b) \(x^2y^2+1-x^2-y^2\)
\(=x^2\left(y^2-1\right)-\left(y^2-1\right)\)
\(=\left(x^2-1\right)\left(y^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
Câu c) đề sai
Câu c) ,đề đúng nek
\(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left[\left(b+c\right)-\left(a+b\right)\right]-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left(b+c\right)-ac\left(a+b\right)-ab\left(a+b\right)\)
\(=\left(b+c\right)\left(bc+ac\right)-\left(a+b\right)\left(ac+ab\right)\)
\(=\left(b+c\right)c\left(a+b\right)-\left(a+b\right)a\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\left(c-a\right)\)
1.a^3-7a-6
<=>x^3+2x^2-2x^2-4x-3x-6
<=>x^2-2x-3(x+2)=(x^2+x-3x-3)(x+2)
<=>[(x-3)(x+1)](x+2)
<=>(x-3)(x+1)(x+2)=0
<=>x-3=0 <=>x=3 hoặc x+1=0<=>x=-1 hoặc x+2=0<=>x=-2
2. a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc
=a(b^2+2bc+c^2)+b(c^2+2ca+a^2)+c(a^2+2ab+b^2)-4abc
=ab^2+2abc+ac^2+bc^2+2abc+ba^2+ca^2+2abc+b^2-4abc
=ab^2+bc^2+ca^2+cb^2+6abc-4abc
=ab^2+bc^2+ca^2+cb^2+2abc
=a^3+b^3+c^3+2abc
1)
b) \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x^2-2xz+z^2\right)-\left(y-1\right)^2\)
\(=\left(y-z\right)^2-\left(y-1\right)^2\)
\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)
\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)
\(A=a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]+4abc\)
\(=a\left(b-c+a\right)\left(b-c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)+4abc\)
\(=\left(a+b-c\right)\left(ab-ac-a^2-bc+ab-b^2\right)+c\left(a^2-2ab+b^2-c^2+4ab\right)\)
\(=\left(a+b-c\right)\left[-c\left(a+b\right)-\left(a-b\right)^2\right]+c\left[\left(a+b\right)^2-c^2\right]\)
\(=\left(a+b-c\right)\left(-ca-cb-a^2+2ab-b^2+ac+cb+c^2\right)\)
\(=\left(a+b-c\right)\left(c^2-\left(a-b\right)^2\right)\)
\(=\left(a+b-c\right)\left(c+a-b\right)\left(a+b-c\right)\)