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\(\text{a) }3x^2+x-2\)
\(=3x^2+3x-2x-2\)
\(=\left(3x^2+3x\right)-\left(2x+2\right)\)
\(=3x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-2\right)\)
b) \(5x^2-18x-8\)
\(=5x^2-20x+2x-8\)
\(=5x\left(x-4\right)+2\left(x-4\right)\)
\(=\left(5x+2\right)\left(x-4\right)\)
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 )
a) \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
a) \(^{x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)}\)
b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)
c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)
\(=-5\left(4z^2+2xy-y-x^2\right)\)
d)\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3xy^2+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a, x4 + 2x3 + x2 = \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2=\left[x\left(x+1\right)\right]^2=\)\(\left(x^2+x\right)^2\)
b, x^3 - x + 3x^2y + 3xy^2+y^3-y
x^3 + 3x^2y + 3xy^2+y^3- x - y
(x+y)^3 - (x+y)
=(x+y)[ (x+y)^2 - 1]
=(x+y)(x+y+1)(x+y-1)
c, 5x^2 - 10xy + 5y^2 - 20(c hỗ này có dấu gì ko???) z^2
\(a,3x^2+x-2\)
\(=3x^2+3x-2x-2\)
\(=\left(x+1\right)\left(3x-2\right)\)
\(b,5x^2-18x-8\)
\(=5x^2-20x+2x-8\)
\(=5x\left(x-4\right)+2\left(x-4\right)\)
\(=\left(x-4\right)\left(5x+2\right)\)
\(c,2x^2-3xy-2y^2\)
\(=2x^2-4xy+xy-2y^2\)
\(=2x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(2x+y\right)\)