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b) \(-y^8+10y^4x^3-25x^6\)
\(=-\left(y^8-10y^4x^3+25x^6\right)\)
\(=-\left[\left(y^4\right)^2-2.y^4.5x^3+\left(5x^3\right)^2\right]\)
\(=-\left(y^4-5x^3\right)^2\)
c) \(8x^3+36x^2y+54xy^2+27y^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^3\)
d) \(-y^3+12y^2x-48yx^2+64x^3\)
\(=-\left(y^3-12y^2x+48yx^2-64x^3\right)\)
\(=-\left[y^3-3.y^2.4x+3.y.\left(4x\right)^2-\left(4x\right)^3\right]\)
\(=-\left(y-4x\right)^3\)
e) \(64x^6y^4-81x^2y^2\)
\(=\left(8x^3y^2\right)^2-\left(9xy\right)^2\)
\(=\left(8x^3y^2-9xy\right)\left(8x^3y^2+9xy\right)\)
f) \(64x^6-27y^6\)
\(=\left(4x^2\right)^3-\left(3y^2\right)^3\)
\(=\left(4x^2-3y^2\right)\left[\left(4x^2\right)^2+4x^2.3y^2+\left(3y^2\right)^2\right]\)
\(=\left(4x^2-3y^2\right)\left(16x^4+12x^2y^2+9x^4\right)\)
a) 10x(x - y)2 - 5(x - y)3 = [10x - 5(x - y)](x - y)2 = (10x - 5x + y)(x - y)2 = (5x + y)(x - y)2
b) -x2 - 10x - 25 = -(x2 + 10x + 52) = -(x + 5)2
c) 64x6y4 - 81x2y2 = (8x3y2)2 - (9xy)2 = (8x3y2 + 9xy)(8x3y2 - 9xy)
d) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)
e)1/8x3 - 3/4x2y + 3/2xy2 - y3 = (1/2x)3 - 3.(1/2x)2y + 3.1/2xy2 - y3 = (1/2x - y)3
f) (3x + 1)2 - (x - 1)2 = (3x + 1 + x - 1)(3x + 1 - x + 1) = 4x(2x + 2) = 8x(x + 1)
a)x^3-x+3x^2y+3xy^2+y^3-y
=(x^3+3x^2y+3xy^2+y^3)-(x+y)
=(x+y)^3-(x+y)
=(x+y)[(x+y)^2-1]
=(x+y)(x+y-1)(x+y+1)
b)81x^2-6yz-9y^2-z^2
=81x^2-(9y^2-6yz+z^2)
=81x^2-(3y-z)^2
=(9x)^2-(3y-z)^2
=(9x-3y+z)(9x+3y-z)
c)12x^2-72x+60
=12(x^2-6x+5)
=12(x^2-x-5x+5)
=12[x(x-1)-5(x-1)]
=12(x-1)(x-5)
y4 + 64 = y4 + 16y2 + 64 - 16y2
<=>y4-y4-16y2+16y2+64-64
<=>0=0
Vậy có vô số y thoa mãn
y4 + 64 = y4 + 16y2 + 64 - 16y2
y4 + 64 = y4 + 16y2 + 64 - 16y2
= (y2 + 8)2 - (4y)2
= (y2 + 8 - 4y)(y2 + 8 + 4y)
a , \(81x^2y+18xy^2+27x^2y^2\)\(=9xy\left(9x+2y+3xy\right)\)
b. \(4x^3+x^2+x=x\left(4x^2+x+1\right)\)
c. \(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\)\(=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
d.
e. \(\left(x+y\right)^3-\left(x-3\right)^3\)\(=x^3+3x^2y+3xy^2+y^3-x^3+9x^2-27x-y^3\)
\(=3x^2y+3xy^2+9x^2-27x\)
\(=3x\left(xy+y^2+3x-9\right)\)
h. \(x^2+x+\frac{1}{4}=\)\(4x^2+4x+1=\left(2x+1\right)^2=\left(2x+1\right)\left(2x+1\right)\)
i.
A = x2(x - 1) + 6(1 - x)
A = x3 - x2 + 6 - 6x
A = (x3 - 6x) - (x2 - 6)
A = x.(x2 - 6) - (x2 - 6)
A = (x - 1)(x2 - 6)
C = x2 + 2xy + y2 - yz - xz
C = (x + y)2 - z.(x + y)
C = (x + y - z).(x + y)
a) Ta có: \(3x^2-6xy+3y^2\)
\(=3\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)^2\)
b) Ta có: \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
c) Ta có: \(64xy-96x^2y+48x^3y-8x^4y\)
\(=8xy\left(8-12x+6x^2-x^3\right)\)
\(=8xy\left(2-x\right)^3\)
d) Ta có: \(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
1) 3-81x3
=3-3.(3x)3
=3[13-(3x)3)]
=3(1-3x)(1+3x+9x2)
2) 250x3y3-2
=2(125x3y3-1)
=2[(5xy)3-13]
=2(5xy-1)(25x2y2+5xy+1)
3) a4+4b4
=[(a2)2+4a2b2+(2b2)]-4a2b2
=(a2+2b2)-(2ab)2
=(a2+2b2-2ab)(a2+2b2+2ab)
4) x5+x+1
=x5-x2+x2+x+1
=x2(x3-13)+(x2+x+1)
=x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x-1)+1]
=(x2+x+1)(x3-x2+1)
Chúc bn học giỏi nhoa!!!