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a) \(ab-ac-b^2+2bc-c^2\)
\(=\left(ab-ac\right)-\left(b^2-2bc+c^2\right)\)
\(=a\left(b-c\right)-\left(b-c\right)^2\)
\(=\left(a-b+c\right)\left(b-c\right)\)
b) \(x^6+8=\left(x^2\right)^3+2^3\)
\(=\left(x^2+2\right)\left(x^4-2x^2+4\right)\)
c) \(64x^3-8=\left(4x\right)^3-2^3\)
\(=\left(4x-2\right)\left(16x^2+8x+4\right)\)
\(=8\left(2x-1\right)\left(4x^2+2x+1\right)\)
d) \(x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x^2+4\right)\left(x-2\right)\)
b: \(=ab^2+ac^2+abc+bc^2+ba^2+abc+a^2c+b^2c+abc\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)
a: \(=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=-x^2\left(y-1\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left(-x^2y-x^2+y+x\right)\)
\(=\left(1-y\right)\left(x^2y+x^2-x-y\right)\)
\(=\left(1-y\right)\cdot\left[y\left(x-1\right)\left(x+1\right)+x\left(x-1\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)
a)=(a2+2ab+b2) +(b2-c2) +(ab+ac)-c2
=(a+b)2 -c2 +(b+c)(b-c) +a(b+c)
=(a+b-c)(a+b+c)+(b+c)(a+b-c)
=(a+b-c)(a+2b+2c)
c)a4+2a3+1
=a4 +a3+a3+a2-a2-a+a+1
=a3(a+1)+a2(a+1)-a(a+1)+(a+1)
=(a+1)(a3+a2-a+1)
d)x5+x+1
=(x5+x4+x3)-x4-x3-x2+x2+x+1
=x3(x2+x+1) -x2(x2+x+1) +(x2+x+1)
=(x2+x+1)((x3-x2+1)
e)x8+x4+1
=(x4)2 +2x4+1-x4
=(x4+1)2 -x4
=(x4+1+x2)(x4+1-x2)
=(x4+2x2+1-x2)(x4-x2+1)
=[(x2+1)2-x2 ](x4-x2+1)
=(x2+1-x)(x2+1 )(x4-x2+1)
phân tích đa thức thành nhân tử
a/x2(x+1)-2x(x+1)+(x+1)=(x+1)(x^2-2x+1)=(x+1)(x-1)^2
b/a2+b2+2a-2b-2ab=(a^2-ab)+(b^2-ab)+2(a-b)=a(a-b)-b(a-b)+2(a-b)=(a-b)(a-b+2)
c/ 4x2-8x+3=(2x-2)^2-1=(2x-2-1)(2x-2+1)=(2x-3)(2x-1)
d/25-16x2=5^2-(4x)^2=(5-4x)(5+4x)
h) (x+1)(x+4)(x+2)(x+3) - 24
= (x2+4x+x+4)(x2+3x+2x+6)-24
=(x2+5x+5-1)(x2+5x+5+1)-24
=(x2+5x+5)2 -12 -24
=(x2+5x+5)2 -25
=(x2+5x+5)2 -52
=(x2+5x+5-5)(x2+5x+5+5)
=(x2+5x)(x2+5x+10)
i) 4(x2+5x+10x+50)(x2+6x+12x+72)-3x2
=4[x(x+5)+10(x+5)].[x(x+6)+12(x+6)]- 3x2
=4(x+10)(x+5)(x+12)(x+6)-3x2
=4(x+10)(x+6)(x+12)(x+5)-3x2
=4(x2+6x+10x+60)(x2+5x+12x+60)-3x2
=4(x2+16x+60)(x2+17x+60)-3x2
Đặt (x2+16x+60) = a
Ta có: 4a(a+x)-3x2
=4a2+4ax -3x2
=(2a)2 + 2.2a.x +x2 -4x2
= [ (2a) +x]2 - (2x)2
= [ (2a) +x -2x].[(2a) + x +2x)]
=[ (2a) -x].[(2a) + 3x)]
sau đó ta thế a = (x2+16x+60) rồi rút gọn là xong ^^
\(A=x^3+4x^2-8x-8=\left(x^3-8\right)+4x\left(x-2\right)=\left(x^3-2^3\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)+4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+4x\right)=\left(x-2\right)\left(x^2+6x+4\right)\)
\(B=a^2+b^2-a^2b^2+ab-a-b=\left(ab-a\right)-\left(a^2b^2-a^2\right)+\left(b^2-b\right)\)
\(=a\left(b-1\right)-a^2\left(b^2-1\right)+b\left(b-1\right)=a\left(b-1\right)-a^2\left(b-1\right)\left(b+1\right)+b\left(b-1\right)\)
\(=\left(b-1\right)\left(a-a^2b-a^2+b\right)\)
\(C=x^4-x^3-x+1=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
Đoàn Thị Huyền Đoan: Hình như câu A bạn chép xuống bị sai đề rồi!
b: \(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
d: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-8\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-8\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-2\right)\)