Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4-2\right)^2-x^4\)
\(=\left(x^4-x^2-2\right)\left(x^4-x^2-2x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+1\right)\left(x^2-1\right)\left(x^2+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-2\right)\left(x^2+1\right)\left(x^2+2\right)\)
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
\(x^8+3x^4+1\)
\(=\left(x^4\right)^2+2x^4.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{5}{4}\)
\(=\left(x^4+\frac{3}{2}\right)^2-\left(\sqrt{\frac{5}{4}}\right)^2\)
\(=\left(x^4+\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\left(x^4+\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\)
Nhận lời thách đố
\(=x^8+\frac{6}{2}x^4+1\)
\(=x^8+\frac{3+\sqrt{5}+3-\sqrt{5}}{2}x^4+\frac{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}\)
\(=x^8+\frac{x^4.\left(3+\sqrt{5}\right)}{2}+\frac{x^4\left(3-\sqrt{5}\right)}{2}+\left(\frac{3+\sqrt{5}}{2}\right)\left(\frac{3-\sqrt{5}}{2}\right)\)
\(=x^4\left(x^4+\frac{3+\sqrt{5}}{2}\right)+\frac{3-\sqrt{5}}{2}\left(x^4+\frac{3+\sqrt{5}}{2}\right)\)
\(=\left(x^4+\frac{3-\sqrt{5}}{2}\right)\left(x^4+\frac{3+\sqrt{5}}{2}\right)\)
Nếu thấy đúng nhớ tk nha
\(x^8y^8+x^4y^4+1=\left[\left(x^4y^4\right)^2+2x^4y^4+1\right]-x^4y^4=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1-x^2y^2\right)\left(x^4y^4+1+x^2y^2\right)\)
\(=\left(x^4y^4+1-x^2y^2\right)\left[\left(x^2y^2\right)^2+2x^2y^2+1-x^2y^2\right]\)
\(=\left(x^4y^4+1-x^2y^2\right)\left[\left(x^2y^2+1\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^4y^4+1-x^2y^2\right)\left(x^2y^2+1-xy\right)\left(x^2y^2+1+xy\right)\)
Phân tích đa thức thành nhân tử
x3+3x2y−9xy2+5y2
x8y8+x4y4+1
\(x^4+3x^2-4\)
\(=x^4+4x^2-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
Chúc bạn học tốt.
x\(x^4+3x^4+4=\left(x^2\right)^2+2x^2\times\frac{3}{2}+\frac{9}{4}\)
Mình xin lỗi nhé, để mình sửa lại : ^^
a) \(x^4+3x^2+4=\left(x^4+x^3+2x^2\right)+-\left(x^3+x^2+2x\right)+2\left(x^2+2x+2\right)\)
\(=x^2\left(x^2+x+2\right)-x\left(x^2+x+2\right)+2\left(x^2+x+2\right)=\left(x^2-x+2\right)\left(x^2+x+2\right)\)
b) \(x^4+5x^2+9=\left(x^4+x^3+3x^2\right)-\left(x^3+x^2+3x\right)+3\left(x^2+x+3\right)\)
\(=x^2\left(x^2+x+3\right)-x\left(x^2+x+3\right)+3\left(x^2+x+3\right)=\left(x^2-x+3\right)\left(x^2+x+3\right)\)
câu b tớ thêm chút
a) x8+3x4+4
=x8-x4+4x4+4
=(x4-1)(x4+1)+4.(x4+1)
=(x4+1)(x4-1+4)
=(x4+1)(x4+3)
b) x6-x4-2x3+2x2
=x4.(x2-1)-2x2.(x-1)
=x4.(x-1)(x+1)-2x2(x-1)
=x2.(x-1)[x2(x+1)-2]
=x2.(x-1)(x3+x2-2)
=x2.(x-1)(x3-1+x2-1)
=x2.(x-1)[(x-1)(x2+x+1)+(x-1)(x+1)]
=x2.(x-1)(x-1)(x2+x+1+x+1)
=x2.(x-1)2.(x2+2x+2)
x8+3x4+4 =x8+4x4+4-x4
=(x4-2)2-x4
=(x4-x2-2)(x4+x2-2)
=(x4-2x2+x2-2)(x4-x2+2x2-2)
=(x2-2)(x2+1)(x2-1)(x2+2)
=(x-1)(x+1)(x2-2)(x2+1)(x2+2)