Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1
Ta có lược đồ sau :
| 1 | 1 | -4 | -4 | |
| -1 | 1 | 0 | -4 | 0 |
Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)
Hai câu đầu tham khảo
Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath
c) \(E=\left(x+a\right)\left(x+2a\right)\left(a+3a\right)\left(x+4a\right)+a^4\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(a+3a\right)+a^4\)
\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)
Đặt \(x^2+5ax+4a^2=t\)
\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)
\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)
Mà \(x^2+5ax+4a^2=t\)
Nên \(\left(2\right)=\left(x^2+5ax+5a^2\right)^2\)
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=\left(x+y+z\right)\left(xy+xz\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z+x\right)\)
\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)
P/s: Sai sót xin bỏ qua.
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz.\)
\(=x^2.\left(y+z\right)+yz.\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)
\(=\left(y+z\right).\left(x^2+yz\right)+x\left(y^{^2}+z^2+2yz\right)\)
\(=\left(y+z\right).\left[x.\left(x+2\right)+y.\left(x+2\right)\right]\)
\(=\left(y+z\right).\left(x+z\right).\left(x+y\right)\)
\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)
a: Ta có: \(25x^2-4a^2+12ab-9b^2\)
\(=25x^2-\left(2a-3b\right)^2\)
\(=\left(5x-2a+3b\right)\left(5x+2a-3b\right)\)
b: Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
\(a^4+a^3+a^{3b}+a^{2b}\)
\(=a\left(a^3+a^2+1^{3b}+1^{2b}\right)\)
\(a^3+3a^2+4a+12\)
\(=a^2\left(a+3\right)+4\left(a+3\right)\)
\(=\left(a^2+4\right)\left(a+3\right)\)