X4次方 1如何因式分解?

x^2-4(x-1)=x^2-4x+4=(x-2)^2x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x+y)(x-y)(5a^2+2a)-4(2+2a^2)=5a^2+2a

X^8+X^7+1=X^8+X^7+X^6-X^6+1=X^6（X^2+X+1）-（X^6-1）=X^6（X^2+X+1）-（X^3+1）（X^3-1）=X^6（X^2+X+1）-（X^3+1）（X-

x^8+1=x^8+1+2x^4-2x^4=(x^4+1)²-2x^4=（x^4+1+√2x²）（x^4+1-√2x²）=（x^4+√2x²+1）（x^4-√2

原式=(x²-y²)²=(x+y)²(x-y)²

3.(1)5x^4+3x^2y-10-3x^2y+x^4-1=6x-11(2)p^2+3pq+6-8p^2+pq=-7p^2+4pq+6(3)(7y-3z)-(8y-5z)=-y+2z(4)-(a^5

x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=x^6(x^2+x+1)+x^3(x^2+x+1)+(x^2+x+1)=(x^6+x^3+1)(x^2+x+1)再问：我也做到了这一步，

LS你没有写对(X+1)(X^4-X^3+X^2+1)=1+X+X^2+X^5楼主的那个题目如果是:x^5-x^4+x^3+x^2+1没有因式得提拉

x^4-4x^3+4x^2-1=x^4-x^3-3x^3+3x^2+x^2-1=x^3(x-1)-3x^2(x-1)+(x+1)(x-1)=(x-1)(x³-3x²+x+1)=(x

令x/1+x2次方=t分母配方得：1+x2次方+x4次方=（x2+1)^2-x^2分子分母同除x^2,得：x2次方/1+x2次方+x4次方=1/{(x2+1)^2/x^2-x^2/x^2}=1/{(1

4分之x4次方y8次方-6x平方y平方+36y4次方=（1/2x²y的4次方-6y²）²如果本题有什么不明白可以追问,

x5-x4+x3-x2+x-1=x4(x-1)+x2(x-1)+(x-1)=(x-1)(x4+x2+1)=(x-1)(x4+2x2+1-x2)=(x-1)[(x2+1)2-x2]=(x-1)(x2+x

1)2x³y—4x平方y平方+2xy³=2xy(x²-2xy+y²)=2xy(x-y)²2）x4次方—2x平方y平方+y4次方=(x²-y&

=x^4+2x^2+1-x^2+2ax-a^2=(x^2+1)^2-(x-a)^2=(x^2+1+x-a)(x^2+1-x+a)

原式=(x^2+1)^2-x^2-2ax-a^2=(x^2+1)^2-(x+a)^2=(x^2+1+x+a）(x^2+1-x-a)x^2意为x的二次方

1.x^4+8x^2-9=x^4-1+8x^2-8=(x^2+1)(x^2-1)+8(x^2-1)=(x+1)(x-1)(x^2+9)2.1/a+1/b=1/61/b+1/c=1/91/a+1/c=1

(x4+1)(x+1)(x-1)再问：过程再答：原式=x4(x2-1)+x2-1=(x4+1)(x2-1)=(x4+1)(x+1)(x-1)再问：已知：a-b=3,.a+c=-5,求ac-bc+a2-

利用a^3-b^3=(a-b)(a^2+ab+b^2)所以x^3-(1/2y)^3=(x-1/2y)(x^2+1/2xy+1/4y^2)

y²-16=(y+4)(y-4)25m²-n²=(5m+n)(5m-n)-x4次方+x²y²=x²(y²-x²)=x&#

x平方-3x+1=0二边同除以xx-3+1/x=0x+1/x=3x^2+1/x^2=(x+1/x)^2-2=3^2-2=7x^4+1/x^4=(x^2+1/x^2)^2-2=7^2-2=47

x^2-4x+1=0两边除以xx+/x=4则x^2+1/x^2=14x^4+1/x^4+2=14^2则x^4+1/x^4=14^2-2