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a,\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)
\(=\left(x-1\right)\left(x\left(x-3\right)+8\left(x-3\right)\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
\(3x^2-5x+2=3x^2-3x-2x+2=3x\left(x-1\right)-2\left(x-1\right)=\left(3x-2\right)\left(x-1\right)\)
b.)x^4+5x^3+15x-9
=x^4-9+5x^3+15x
=(x^2-3)(x^2+3)+5x(x^2+3)
=(x^2+3)(x^2-3+5x)
Ta có: (x^2+2x).(x^2+4x+3)-24
=x.(x+2).(x^2+x+3x+3)-24
=x.(x+2).[x.(x+1)+3.(x+1)]-24
=x.(x+2).(x+1).(x+3)-24
=[x.(x+3)].[(x+1).(x+2)]-24
=(x^2+3x).(x^2+3x+2)-24
Đặt x^2+3x=t,phương trình trở thành: t.(t+2)-24=t^2+2t-24=t^2-4t+6t-24=t.(t-4)+6.(t-4)=(t-4).(t+6)=(x^2+3x-4).(x^2+3x+6)
a) 4x3 + 4x2 + 6x
= 2x( 2x2 + 2x + 3 )
b) x3 + 4x2 - 29x + 24
= x3 + 8x2 - 4x2 - 32x + 3x + 24
= ( x3 - 4x2 + 3x ) + ( 8x2 - 32x + 24 )
= x( x2 - 4x + 3 ) + 8( x2 - 4x + 3 )
= ( x2 - 4x + 3 )( x + 8 )
= ( x2 - x - 3x + 3 )( x + 8 )
= [ x( x - 1 ) - 3( x - 1 ) ]( x + 8 )
= ( x - 1 )( x - 3 )( x + 8 )
c) x3 + 21x2 + 134x + 240
= x3 + 10x2 + 11x2 + 110x + 24x + 240
= ( x3 + 11x2 + 24x ) + ( 10x2 + 110x + 240 )
= x( x2 + 11x + 24 ) + 10( x2 + 11x + 24 )
= ( x + 10 )( x2 + 11x + 24 )
= ( x + 10 )( x2 + 3x + 8x + 24 )
= ( x + 10 )[ x( x + 3 ) + 8( x + 3 ) ]
= ( x + 10 )( x + 8 )( x + 3 )
d) 25x3 - 25x2y - x + y
= 25x2( x - y ) - 1( x - y )
= ( x - y )( 25x2 - 1 )
= ( x - y )( 5x - 1 )( 5x + 1 )
e) 2x5y - 4x3y + 2xy
= 2xy( x4 - 2x2 + 1 )
Đặt t = x2
= 2xy( t2 - 2t + 1 )
= 2xy( t - 1 )2
= 2xy( x2 - 1 )2
= 2xy[ ( x - 1 )( x + 1 ) ]2
= 2xy( x - 1 )2( x + 1 )2
g) x3 - 3x + 2
= x3 + 2x2 - 2x2 - 4x + x + 2
= ( x3 - 2x2 + x ) + ( 2x2 - 4x + 2 )
= x( x2 - 2x + 1 ) + 2( x2 - 2x + 1 )
= ( x + 2 )( x2 - 2x + 1 )
= ( x + 2 )( x - 1 )2
\(\left(x^2+4x-3\right)^2-5x.\left(x^2+4x-3\right)+6x^2\)
\(=\left[\left(x^2+4x-3\right)^2-2.\left(x^2+4x-3\right).2,5x+\left(2,5x\right)^2\right]-\left(0,5x\right)^2\)
\(=\left(x^2+4x-3-2,5x\right)^2-\left(0,5x\right)^2\)
\(=\left(x^2+4x-3-2,5x-0,5x\right).\left(x^2-4x-3-2,5x+0,5x\right)\)
\(=\left(x^2+x-3\right).\left(x^2+2x-3\right)\)
Tham khảo nhé~
4x3+4x4−x2−x
=4x3(x+1)−x(x+1)
=(x+1)(4x3−1)
ĐÂY NHÉ. T.I.C.K MÌNH VỚI