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Phân tích đa thức thành nhân tử:
1. x2 - 5x - 6
\(=x^2-6x+x-6\)
\(=\left(x^2-6x\right)+\left(x-6\right)\)
\(=x\left(x-6\right)+\left(x-6\right)\)
\(=\left(x+1\right)\left(x-6\right)\)
3. 1 + 3x + 2x2
\(=1+2x+x+2x^2\)
\(=\left(1+2x\right)+\left(x+2x^2\right)\)
\(=\left(1+2x\right)+x\left(1+2x\right)\)
\(=\left(1+2x\right)+\left(x+1\right)\)
4. 6 - 2x - 8x2
\(=6-8x+6x-8x^2\)
\(=\left(6+6x\right)-\left(8x+8x^2\right)\)
\(=6\left(1+x\right)-8x\left(1+x\right)\)
\(=\left(6-8x\right)\left(1+x\right)\)
5. 7 - 4x - 3x2
\(=7-7x+3x-3x^2\)
\(=\left(7-7x\right)+\left(3x-3x^2\right)\)
\(=7\left(1-x\right)+3x\left(1-x\right)\)
\(=\left(7+3x\right)\left(1-x\right)\)
1) \(x^2-5x-6\\ =x^2+x-6x-6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x-6\right)\left(x+1\right)\)
3) \(1+3x+2x^2\\ =2x^2+2x+x+1\\ =2x\left(x+1\right)+\left(x+1\right)\\ =\left(x+1\right)\left(2x+1\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
1)x2-3x+2
=x2-x-2x+2
=x(x-1)-2(x-1)
=(x-1)(x-2)
2)2x2+5x-7
=2x2-2x+7x-7
=2x(x-1)+7(x-1)
=(x-1)(2x+7)
3)x2+7x-8
=x2-x+8x-8
=x(x-1)+8(x-1)
=(x-1)(x+8)
4)x2+8x+7
=x2+x+7x+7
=x(x+1)+7(x+1)
=(x+1)(x+7)
5)2x2-5x-7
=2x2+2x-7x-7
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
xong rồi !!
1) 4x2 + 5x - 6 = 4x2 + 8x - 3x - 6 = 4x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( 4x - 3 )
2) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
3) 2x2 + 3x - 27 = 2x2 - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 ) < đã sửa ._. >
4) 7x2 + 3xy - 10y2 = 7x2 - 7xy + 10xy - 10y2 = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )
5) x2 + 5x - 2 < sai đề ._. >
6) x8 + x7 + 1 = x8 + x7 + x6 - x6 + 1
= ( x8 + x7 + x6 ) - ( x6 - 1 )
= x6( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x6( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )[ x6 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
1. \(x^2+3x+2=x^2+2x+x+2=x\left(x+2\right)+\left(x+2\right)=\left(x+2\right)\left(x+1\right)\)
2.
\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(3x-1\right)\)3 và 4 bạn xem lại đề nha mik giải ko đc
5. \(x^2+2x-8=x^2+4x-2x-8=x\left(x+4\right)-2\left(x+4\right)=\left(x+4\right)\left(x-2\right)\)
cảm ơn bn nhiều nhé!!!!