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(2x2 - y)3 - 64y3 = (2x2 - y - 4y) [ (2x2 - y)2 + (2x2 - y).4y + 16y2 ]
= (2x2 - 5y) (4x2 - 4x2y + y2 + 8x2y - 4y2 + 16y2)
= (2x2 - 5y) (4x2 + 4x2y + 13y2)
\(1.=\left(2-3a-3+a\right)\left(2-3a+3-a\right)\)
\(=\left(-1-2a\right)\left(5-4a\right)\)
\(2.=x\left(x^2-1\right)+2\left(x^2-1\right)\)
\(=\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
a/ \(=64y^4+32xy^3+8y^2x^2-32xy^3-16x^2y^2-4x^3y+8x^2y^2+4x^3y+x^4\)
\(=8y^2\left(8y^2+4xy+x^2\right)-4xy\left(8y^2+4xy+x^2\right)+x^2\left(8y^2+4xy+x^2\right)\)
\(=\left(8y^2-4xy+x^2\right)\left(8y^2+4xy+x^2\right)\)
b/ \(=y^4+2xy^3+2x^2y^2-2xy^3-4x^2y^2-4x^3y+2x^2y^2+4x^3y+4x^4\)
\(=y^2\left(y^2+2xy+2x^2\right)-2xy\left(y^2+2xy+2x^2\right)+2x^2\left(y^2+2xy+2x^2\right)\)
\(=\left(y^2-2xy+2x^2\right)\left(y^2+2xy+2x^2\right)\)
c/ \(=x^4+5x^3+7x^2+5x^3+25x^2+35x+3x^2+15x+21\)
\(=x^2\left(x^2+5x+7\right)+5x\left(x^2+5x+7\right)+3\left(x^2+5x+7\right)\)
\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)
d/ \(=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
a) 9 + 10ab - a2 - 25b2 = 9 - (a2 - 10ab + 25b2)
= 9 - (a - 5b)2
= (3 - a + 5b)(3 + a - 5b)
b) 16x2 - (x2 - 2x + 1) = 16x2 - (x - 1)2
= (4x - x + 1)(4x + x - 1)
= (3x + 1)(5x - 1)
d) 12xy2 + 27y2 - 3y3 - 12x2 = 3(4xy2 + 9y2 - y3 - 4x2)
f) x4 + 64y4 = x4 + 16x2y2 + 64y4 - 16x2y2
= (x2 + 8y2)2 - 16x2y2
= (x2 - 4xy + 8y2)(x2 + 4xy + 8y2)
a,\(0,04x^2-64y^2=\left(0,2x\right)^2-\left(8y\right)^2=\left(0,2x-8y\right)\left(0,2x+8y\right)\)
b,\(-x^3+9x^2-27x+27=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)=\left(-x^2+6x-9\right)\left(x-3\right)\)
\(=-\left(x-3\right)^3\)
Nhớ tick mình nha bạn,cảm ơn nhiều nha.
a) 2x2-6x-x+3 = 2x(x-3) - (x-3) = (x-3)(2x-1)
b) x2-x-5x+5 = x(x-1) - 5(x-1) = (x-1)(x-5)
c) 5x(x-2y) + 2( x-2y)2 = (x-2y)(5x+2x-2y) = (x-2y)(7x-2y)
chú ý : (A-B)2=(B-A)2
d) 7x(4-y)2 - (4-y)3 = ( 16-8y+y2) (7x-4+y)
a) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(2x-1\right)\)
b) \(x^2-6x+5=x^2-5x-x+5=x\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x-1\right)\)
c)\(5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
d) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)+\left(y-4\right)^3=\left(y-4\right)\left(7x-y-4\right)\)
1)\(x^3-2x^2y+x-xz^2\)
\(=x\left(x^2-2xy+1-z^2\right)\)
\(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
2) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\left(x-2\right)\left[x^2+2x+7+2.\left(x+2\right)-5\right]=0\)
\(\left(x-2\right)\left(x^2+6\right)=0\)
Ta có: \(x^2+6>0\forall x\)
Để \(\left(x-2\right)\left(x^2+6\right)=0\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\left(x-2\right)^2=\left(3x-2\right)\left(x-2\right)\)
\(\left(x-2\right)^2-\left(3x-2\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x-2-3x+2\right)=0\)
\(-2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-2x=0\\x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
P/S: Câu 1 nghi sai đề
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( 2x2 - y )3 - 64y3
= ( 2x2 - y )3 - ( 4y )3
= ( 2x2 - y - 4y )[ ( 2x2 - y )2 + ( 2x2 - y ).4y + ( 4y )2 ]
= ( 2x2 - 5y )( 4x4 - 4x2y + y2 + 8x2y - 4y2 + 16y2 )
= ( 2x2 - 5y )( 4x4 + 4x2y + 13y2 )