Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x6+x4+x2y2+y4-y6=(x6-y6)+(x4+x2y2+y4)=(x2-y2)(x4+x2y2+y4)+(x4+x2y2+y4)=(x4+x2y2+y4)(x2-y2+1)=((x2+y2)2-x2y2)(x2-y2+1)
=(x2+xy+y2)(x2-xy+y2)(x2-y2+1)
x4-30x2+31x-30=(x4+x)-(30x2-30x+30)=x(x+1)(x2-x+1)-30(x2-x+1)=(x2-x+1)(x2+x-30)=(x2-x+1)(x-5)(x+6)
\(x^4+y^2-2x^2y+x^2+2x-2y\)
\(=\left(y^2-x^2y-xy\right)-\left(x^2y-x^4-x^3\right)+\left(xy-x^3-x^2\right)-\left(2y-2x^2-2x\right)\)
\(=y\left(y-x^2-x\right)-x^2\left(y-x^2-x\right)+x\left(y-x^2-x\right)-2\left(y-x^2-x\right)\)
\(=\left(y-x^2+x-2\right)\left(y-x^2-x\right)\)
\(x^4-2x^2y^2+y^4-1=0\Leftrightarrow\left(x^2-y^2\right)^2-1=0\Leftrightarrow\left(x^2-y^2-1\right).\left(x^2-y^2+1\right)=0\\ \)
\(x^2+2xy+2x+2y+y^2+1=0\Leftrightarrow\left(x+y+1\right)^2=0\)
a)\(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
b)\(x^2.\left(1-x^2\right)-4+4x^2=x^2.\left(1-x^2\right)-4.\left(1-x^2\right)=\left(1-x^2\right).\left(x^2-2^2\right)\)\(=\left(1-x\right).\left(1+x\right).\left(x-2\right).\left(x+2\right)\)
Tham khảo nhé~
Bài giải:
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
\(x^2+5x-6\)
\(\Leftrightarrow x^2-x+6x-6\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\)
Chúc bạn học tốt
\(x^3-x+3x^2y+xy^2+y^3-y\)
\(=\left(x^3+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(x^2+y^2-x^2y^2+xy-x-y\)
\(\Leftrightarrow x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)\)
\(\Leftrightarrow\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(\Leftrightarrow\left(1-y\right)\left(x+y\right)\left(x-1\right)\left(x+1\right)\)
c)Ta có:
\(x^8+x+1=x^8-x^2+x^2+x+1=\left(x^8-x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^6-1\right)+\left(x^2+x+1\right)=x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
phân tích đa thức thành nhân tử
a.x3y3+x2y2+4
b.x4+y4+(x+y)4
c.x8+x+1
=> x = ............
\(x^4+x^2y^2+y^2\)
\(=x^4+2x^2y^2-x^2y^2+y^2\)
\(=\left(x^4+2x^2y^2+y^2\right)-x^2y^2\)
\(=\left(x^2+y\right)^2-x^2y^2\)
\(=\left(x^2+y-x^2y^2\right)\left(x^2+y+x^2y^2\right)\)