Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=4y^2-\left(x^2-10x+25\right)\)
\(A=4y^2-\left(x-5\right)^2\)
\(A=\left(2y-x-5\right)\left(2y+x-5\right)\)
\(B=\left(x-4\right)^4-\left(x+a\right)^4\)
\(B=\left(\left(x-4\right)^2\right)^2-\left(\left(x+a\right)^2\right)^2\)
\(B=\left(\left(x-4\right)^2-\left(x+a\right)^2\right)\left(\left(x-4\right)^2+\left(x+a\right)^2\right)\)
\(B=\left(x-4\right)\left(x+a\right)\left(\left(x-4\right)^2+\left(x+a\right)^2\right)\)
\(C=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(C=\left(x^2+x\right)\left(x^2+x+2\right)+1\)
Bài 1:
a)x2-10x+9
=x2-x-9x+9
=x(x-1)-9(x-1)
=(x-9)(x-1)
b)x2-2x-15
=x2+3x-5x-15
=x(x+3)-5(x+3)
=(x-5)(x+3)
c)3x2-7x+2
=3x2-x-6x+2
=x(3x-1)-2(3x-1)
=(x-2)(3x-1)x^3-12+x^2
d)x3-12+x2
=x3+3x2+6x-2x2-6x-12
=x(x2+3x+6)-2(x2+3x+6)
=(x-2)(x2+3x+6)
a.\(x^2y-xz+z-y=\)\(\left(x^2y-y\right)-\left(xz-z\right)=\)\(y\left(x^2-1\right)-z\left(x-1\right)\)
\(y\left(x+1\right)\left(x-1\right)-z\left(x-1\right)\)=\(\left(x-1\right)\left(xy+y-z\right)\)
b.\(x^4-x^3+x^2-1=x^3\left(x-1\right)+\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x^3+x+1\right)\)
c.\(x^4-x^2+10x-25=x^4-\left(x^2-10x+25\right)\)=\(\left(x^2\right)^2-\left(x-5\right)^2=\left(x^2+x-5\right)\left(x^2-x+5\right)\)
a,2x2-7x+6=(2x2-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x2+x-6=(x2+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x3+3x2+6x+4=x3+x2+2x2+2x+4x+4
=(x+1)(x2+2x+4)
d,x10+x5+1=(x10-x)+(x5-x2)+(x2+x+1)
=x((x3)3-1)+x2(x3-1)+(x2+x+1)
=x(x3-1)(x6+x3+1)+x2(x-1)(x2+x+1)+(x2+x+1)
=x(x-1)(x2+x+1)+x2(x-1)(x2+x+1)+(x2+x+1)
(x2+x+1)(x2-x+x3-x2+1)
e,(12x2-12xy+3y2)-10x(2x-y)=3(4x2-4xy+y2)-10x(2x-y)
=3(2x-y)2-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)
phân tích đa thức thành nhân tử
a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
Bài 1 :
Tự phân tích vế trái và điền vào vế phải
Bài 2 :
a) \(3x^3-6x^2+3x\)
\(=3x\left(x^2-2x+1\right)\)
\(=3x\left(x-1\right)^2\)
b) \(2xy+z+2x+yz\)
\(=\left(2xy+2x\right)+\left(z+yz\right)\)
\(=2x\left(y+1\right)+z\left(y+1\right)\)
\(=\left(y+1\right)\left(2x+z\right)\)
c) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
d) \(3x^2-4x-7\)
\(=3x^2+3x-7x-7\)
\(=3x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-7\right)\)
d)5.(x-y)-y(x-y)
=(x-y)(5-y)
e) y.(x-z)+7(z-x)
=y.(x-z)-7(x-z)
=(x-z)(y-7)
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
b: \(B=\left(2x-4\right)^2+2\cdot\left(2x-4\right)\left(x+1\right)+\left(x+1\right)^2\)
=(2x-4+x+1)^2
=(3x-3)^2
Khi x=-1/2 thì B=(-3/2-3)^2=(-9/2)^2=81/4
c: \(C=x^2\left(5-4\right)+y^2\left(4-6\right)+z^2\left(6+4\right)\)
=x^2-2y^2+10z^2
=6^2-2*5^2+10*4^2
=146
d: x=9 thì x+1=10
\(D=x^{2017}-x^{2016}\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
=x^2017-x^2017+x^2016+...-x^3-x^2+x^2+x-x-1
=-1
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128