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\(a,x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(b,a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
\(c,8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(d,8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
\(a)x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(b)a^6-b^3=\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(c)8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(d)8z^3+27=\left(2z\right)^3+3^3=\left(2x+3\right)\left(4z^2-6z+9\right)\)
Bạn sai đề rồi :
a ) \(8y^3-125\)
\(=\left(2y\right)^3-5^3\)
\(=\left(2y-5\right)\left(4y^2+2y.5+5^2\right)\)
\(=\left(2y-5\right)\left(4y^2+10y+25\right)\)
b) Ta thấy \(8z^3=\left(2z\right)^3\)còn \(27=3^3\)
Trở thành : \(\left(2z\right)^3+3^3\)
Rồi bạn bạn tự làm nha
Thanks
\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
1. \(x^3+8y^3\)
\(=x^3+\left(2y\right)^3\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)^2\)
\(=\left(x-y\right)^3\)
2. \(a^6-b^3\)
\(=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\left(\left(a^2\right)^2+a^2b+b^2\right)\)
\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
1. x3 + 8 = (x + 2 )(x2 - x + 1)
2. 27 - 8y3 = ( 3 - 2y ) ( 9 + 6y + 4y2 )
3. y6 + 1 = (y2)3 + 1 = ( y2 + 1) ( y4 - y2 +1 )
4.64x3 - \(\dfrac{1}{8}\)y3 = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x2 + 2xy + \(\dfrac{1}{4}\)y2)
5. 125x6 - 27y9 = (5x2)3 - (3y3)3
= ( 5x2 - 3y3)(25x4 +15x2y3 + 9y6)
a) (x+2) \(\left(x^2-2x+4\right)\)
b) (3 - 2y) \(\left(9+6y+4y^2\right)\)
d) (4x - y) \(\left(16x^2+4xy+y^2\right)\)
bạn giải chi tiết hộ mình với nha.
mk sắp phải nộp bài r. huhuhuhu
a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
a ) \(x^3+8y^3=x^3+\left(2y\right)^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b ) \(a^6-b^3=\left(a^2\right)^3-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c ) \(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
d ) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2-6z+9\right)\)
a) x3 + 8y3 = x3 + (2y)3 = (x+2y)(x2+2xy+4y2)
b) a6 - b3 = (a2)3 - b3 = (a2-b)(a4 + a2b + b2)
c) 8y3 - 125 = (2y)3 - 53 = (2y - 5)(4y2 + 10y + 25)
d) 8x3 + 27 = (2z)3 + 33 = (2z + 3)(4z2 - 6x + 9)