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\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
a) x3 + y3 - 3xy + 1
= ( x + y )3 - 3xy( x + y ) - 3xy + 1
= [ ( x + y )3 + 1 ] - [ 3xy( x + y ) + 3xy ]
= ( x + y + 1 )( x2 + 2xy + y2 - x - y + 1 ) - 3xy( x + y + 1 )
= ( x + y + 1 )( x2 - xy + y2 - x - y + 1 )
b) ( 4 - x )5 + ( x - 2 )5 - 32
= [ -( x - 4 ) ]5 + ( x - 2 )5 - 32
Đặt t = x - 3
đthức <=> ( 1 - t )5 + ( 1 + t )5 - 32 ( chỗ này bạn dùng nhị thức Newton để khai triển nhé )
= 10t4 + 20t2 - 30
Đặt y = t2
đthức = 10y2 + 20y - 30
= 10y2 - 10y + 30y - 30
= 10y( y - 1 ) + 30( y - 1 )
= 10( y - 1 )( y + 3 )
= 10( t2 - 1 )( t2 + 3 )
= 10( t - 1 )( t + 1 )( t2 + 3 )
= 10( x - 3 - 1 )( x - 3 + 1 )[ ( x - 3 )2 + 3 ]
= 10( x - 4 )( x - 2 )( x2 - 6x + 12 )
a,\(x^3+y^3-3xy+1\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy\)
\(=\left[\left(x+y\right)^3+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)\)
\(a^7+a^2+1=a^7-a+a^2+a+1=a\left(a^3-1\right)\left(a^3+1\right)+\left(a^2+a+1\right)\)
\(=a\left(a-1\right)\left(a^2+a+1\right)\left(a^3+1\right)+\left(a^2+a+1\right)\)
\(=\left(a^2+a+1\right)\left[a\left(a-1\right)\left(a^3+1\right)+1\right]=\left(a^2+a+1\right)\left(a^5-a^4+a^2-a+1\right)\)
a^5+a+1=a^5-a^2+(a^2+a+1)
=a^2(a^3-1)+(a^2+a+1)
a^2(a-1)(a^2+a+1)+(a^2+a+1)
(a^2+a+1)(a^3-a^2+1)
(a^2+a+1)(
a/ 9a^3 - 13a + 6 = 9a^3 - 6a^2 + 6a^2 - 4a - 9a + 6 = (9a^3 - 6a^2) + (6a^2 - 4a) - (9a - 6) = 3a^2(3a - 2) + 2a(3a - 2) - 3(3a - 2) = (3a^2 + 2a - 3)(3a - 2) Mình gửi luôn cho nóng^^Được câu nào hay câu đó. Yên tâm mình sẽ cố nghĩ &gửi nốt :)))
b/x^4 - 4x^3 + 8x + 3 = x^4 - 3x^3 - x^3 + 3x^2 - 3x^2 + 9x - x + 3 = (x^4 - 3x^3) - (x^3 - 3x^2) - (3x^2 - 9x) - (x - 3) = x^3(x - 3) - x^2(x - 3) - 3x(x - 3) - (x - 3) = (x^3 - x^2 - 3x - 1)(x - 3) Mình đang cố nghĩ nốt con c đây, có vẻ khó^^
a) x3 - 1 + 5x2 - 5 + 3x - 3
= x3 + 5x2 + 3x - 9
= x3 + 6x2 - x2 + 9x - 6x - 9
= ( x3 + 6x2 + 9x ) - ( x2 + 6x + 9 )
= x( x2 + 6x + 9 ) - ( x2 + 6x + 9 )
= ( x2 + 6x + 9 )( x - 1 )
= ( x + 3 )2( x - 1 )
b) a5 + a4 + a3 + a2 + a + 1
= ( a5 + a4 + a3 ) + ( a2 + a + 1 )
= a3( a2 + a + 1 ) + 1( a2 + a + 1 )
= ( a2 + a + 1 )( a3 + 1 )
= ( a2 + a + 1 )( a + 1 )( a2 - a + 1 )
c) x3 - 3x2 + 3x - 1 - y3
= ( x3 - 3x2 + 3x - 1 ) - y3
= ( x - 1 )3 - y3
= ( x - 1 - y )[ ( x - 1 )2 + ( x - 1 )y + y2 ]
= ( x - 1 - y )( x2 - 2x + 1 + xy - y + y2 )
d) 5x3 - 3x2y - 45xy2 + 27y3
= ( 5x3 - 45xy2 ) - ( 3x2y - 27y3 )
= 5x( x2 - 9y2 ) - 3y( x2 - 9y2 )
= ( 5x - 3y )( x2 - 9y2 )
= ( 5x - 3y )[ x2 - ( 3y )2 ]
= ( 5x - 3y )( x - 3y )( x + 3y )
1/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )
\(a^6-a^4+2a^3+2a^2\)
\(=\left[\left(a^3\right)^2-\left(a^2\right)^2\right]+2\left(a^2+a^3\right)\)
\(=\left(a^3-a^2\right)\left(a^3+a^2\right)+2\left(a^3+a^2\right)\)
\(=\left(a^3-a^2+2\right)\left(a^3+a^2\right)\)
\(=a^2.\left(a^3-a^2+2\right)\left(a+1\right)\)
\(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a^2-1\right)+2\left(a+1\right)\right]\)
\(=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
Ta có: \(a^5+a^4+a^3+a^2+a+1=a^3\left(1+a+a^2\right)+1\left(1+a+a^2\right)=\left(a^3+1\right)\left(1+a+a^2\right)\)