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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) 2x3 + 6xy - x2z - 3yz
= ( 2x3 + 6xy ) - ( x2z + 3yz )
= 2x( x2 + 3y ) - z( x2 + 3y )
= ( x2 + 3y )( 2x - z )
b) x2 - 6xy + 9y2 - 49
= ( x2 - 6xy + 9y2 ) - 49
= ( x - 3y )2 - 72
= ( x - 3y - 7 )( x - 3y + 7 )
c) x3 + 4x2 + 16x + 64
= ( x3 + 4x2 ) + ( 16x + 64 )
= x2( x + 4 ) + 16( x + 4 )
= ( x + 4 )( x2 + 16 )
a) =(2x^3-x^2z)+(6xy-3yz)
=x^2(2x-z)+3y(2x-z)
=(x^2+3y)(2x-z)
b) =(x^2-6xy+9y^2)-7^2
=(x-3y)^2-7^2
=(x-3y+7)(x-3y-7)
c) =(x^3+4x^2)+(16x+64)
=x^2(x+4)+16(x+4)
=(x^2+16)(x+4)
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)
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
A= x2-100y2
= x2-(10y)2
= (x-10y)(x+10y)
B= 64x3+1
= (4x)3+13
=(4x+1)(16x2-4x+1)