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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
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)
\(x^3+27y^3=x^3+\left(3y\right)^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(a^6-8b^3=\left(a^2\right)^3-\left(2b\right)^3=\left(a^2-2b\right)\left(a^4+2a^2b+4b^2\right)\)
\(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(8z^3+x^3=\left(2z\right)^3+x^3=\left(2z+x\right)\left(4z^2-2xz+x^2\right)\)
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
a) \(8x^3-27y^6\)
\(=\left(2x\right)^3-\left(3y^2\right)^3\)
\(=\left(2x-3y^2\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)
\(=\left(2x-3y^2\right)\left(4x^2+6xy+9y^2\right)\)
b) \(a^3b^3c^3-1\)
\(=\left(abc\right)^3-1^3\)
\(=\left(abc-1\right)\left(a^2b^2c^2+abc+1\right)\)
c) \(64x^3+\dfrac{1}{8}y^3\)
\(=\left(4x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(4x+\dfrac{1}{2}y\right)\left[\left(4x\right)^2+4x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(4x+\dfrac{1}{2}y\right)\left(4x^2+2xy+\dfrac{1}{4}y^2\right)\)
d) \(125+y^3\)
\(=5^3+y^3\)
\(=\left(5+y\right)\left(25-5y+y^2\right)\)
e) \(a^6-b^6\)
\(=\left(a^3\right)^2-\left(b^3\right)^2\)
\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
f) \(4x^2-9\left(3x+5\right)^2\)
\(=\left(2x\right)^2-\left[3\left(3x+5\right)\right]^2\)
\(=\left[2x-3\left(3x+5\right)\right]\left[2x+3\left(3x+5\right)\right]\)
\(=\left(2x-9x-15\right)\left(2x+9x+15\right)\)
\(=\left(-7x-15\right)\left(11x+15\right)\)
a) 9x4+16y6-24x2y3
=(3x2)2-2.3x2.4y3+(4y3)2
=(3x2-4y3)2
b) 16x2-24xy+9y2
=(4x)2-2.4x.3y+(3y)2
=(4x-3y)2
c) 36x2-(3x-2)2
=(36x-3x+2)(36x+3x-2)
=(33x+2)(39x-2)
d) 27x3+54x2y+36xy2+8y3
=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3
=(3x+2y)3
e) y9-9x2y6+27x4y3-27x6
=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3
=(y3-3x2)3
f) 64x3+1
= (4x)3+13
=(4x+1)[(4x)2-4x.1+12]
=(4x+1)(16x2-4x+1)
e) 27x6-8x3 *sửa đề*
=(3x2)3-(2x)3
=(3x2-2x)[(3x)2+3x2.2x+(2x)2]
=(3x2-2x)(9x2+6x3+4x2)
~~~
D) 64x^3-1/8y^3
= (4x)^3 + (1/2y)^3
= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]
E) 125x^6-27y^9
( câu này mik chưa rõ nên vx chưa tek giải cho bn )
HOk tốt nhé
a.x^3 + 8
= x^3 + 2^3
= ( x+2) ( x^2 - x.2+2^2)
b/27-8y^3
3^3 - (2y)^3
= ( 3 - (2y))(3^2 + 3.2y + (2y)^2)
c/ y^6 + 1
= (y^3)^3 + 1 ^3
= (y^3 + 1)((y^3)^2 - y^3.1 + 1^2