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a,\(2axy-4a^2xy^2+6a^3x^2\)
\(=2ax\left(y-2ay^2+3a^2x\right)\)
b,\(5a^2xy-10a^3x-15ay\)
\(=5a\left(axy-2ax-3y\right)\)
Tự lm tiếp, tất cả đều dùng bằng phương phép đặt nhân tử chung nhé
23) \(2axy-4a^2xy^2+6a^3x^2\)
\(=2ax\left(y-2ay^2+3a^2x\right)\)
24) \(5a^2xy-10a^3x-15ay\)
\(=5a\left(axy-2a^2x-3y\right)\)
25) \(mxy-m^2x+my\)
\(=m\left(xy-mx+y\right)\)
26) \(2mx-4m^2xy+6mx\)
\(=2mx\left(1-2my+3\right)\)
\(=4mx\left(2-my\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
a) \(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+x-3=\left(x-3\right)\left(1-5x\right).\)
b) \(x^4+64=x^4+16x^2+64-16x^2=\left(x^2+8\right)^2-\left(4x\right)^2=\left(x^2+4x+8\right)\left(x^2-4x+8\right).\)
c) \(64x^2+4y^4=4\left(16x^2+y^4\right)\)
d) \(x^5+x-1\)đa thức này có nghiệm vô tỷ. Mik ko phân tích được.
a) \(x^3-x^2-4=x^3-2x^2+x^2-4=x^2\left(x-2\right)+\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
c) \(2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2=2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right).\)
\(=\left(x-2\right)\left(2x^2-8x+1\right)\)
d) \(2x^4+x^3-22x^2+15x+36=2x^4+2x^3-x^3-x^2-21x^2-21x+36x+36.\)
\(=2x^3\left(x+1\right)-x^2\left(x+1\right)-21x\left(x+1\right)+36\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^3-x^2-21x+36\right)\)
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) 10x3-90x5=10x3(1-9x2)=10x3(1-3x)(1+3x)
b)3a2-6a2b+5a-10ab=(3a2-6a2b)+(5a-10ab)=3a2(1-2b)+5a(1-b)=(3a2+5a)(1-2b)=a(3a+5)(1-2b)
c) 7a3-28a2+28a=7a(a2-4a+4)
d) 45a3-30a2+5a-500=5(9a3-6a2+a-100)