B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Bài 1.
a) 5( x2 + y2 ) - 20x2y2
= 5x2 + 5y2 - 20x2y2
= 5( x2 + y2 - 4x2y2 )
b) 2x8 - 32
= 2( x8 - 16 )
= 2[ ( x4 )2 - 42 ]
= 2( x4 - 4 )( x4 + 4 )
= 2[ ( x2 )2 - 22 ]( x4 + 2x3 - 2x3 + 2x2 - 4x2 + 2x2 - 4x + 4x + 4 )
= 2( x2 - 2 )( x2 + 2 )[ ( x4 + 2x3 + 2x2 ) - ( 2x3 + 4x2 + 4x ) + ( 2x2 + 4x + 4 ) ]
= 2( x2 - 2 )( x2 + 2 )[ x2( x2 + 2x + 2 ) - 2x( x2 + 2x + 2 ) + 2( x2 + 2x + 2 ) ]
= 2( x2 - 2 )( x2 + 2 )( x2 + 2x + 2 )( x2 - 2x + 2 )
Bài 2.
a) ( x2 - 3x - 1 )2 - 12( x2 - 3x - 1 ) + 27
= [ ( x2 - 3x - 1 )2 - 12( x2 - 3x - 1 ) + 36 ] - 9
= [ ( x2 - 3x - 1 ) - 6 ) ]2 - 9
= ( x2 - 3x - 7 )2 - 32
= ( x2 - 3x - 7 - 3 )( x2 - 3x - 7 + 4 )
= ( x2 - 3x - 10 )( x2 - 3x - 4 )
= ( x2 + 2x - 5x - 10 )( x2 + x - 4x - 4 )
= [ x( x + 2 ) - 5( x + 2 ) ][ x( x + 1 ) - 4( x + 1 ) ]
= ( x + 2 )( x - 5 )( x + 1 )( x - 4 )
b) ( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24
= [ ( x + 2 )( x + 5 ) ][ ( x + 3 )( x + 4 ) ] - 24
= [ x2 + 7x + 10 ][ x2 + 7x + 12 ] - 24 (*)
Đặt t = x2 + 7x + 10
(*) <=> t( t + 2 ) - 24
= t2 + 2t - 24
= ( t2 + 2t + 1 ) - 25
= ( t + 1 )2 - 25
= ( t + 1 - 5 )( t + 1 + 5 )
= ( t - 4 )( t + 6 )
= ( x2 + 7x + 10 - 4 )( x2 + 7x + 10 + 6 )
= ( x2 + 7x + 6 )( x2 + 7x + 16 )
= ( x2 + x + 6x + 6 )( x2 + 7x + 16 )
= [ x( x + 1 ) + 6( x + 1 ) ]( x2 + 7x + 16 )
= ( x + 1 )( x + 6 )( x2 + 7x + 16 )
Phân tích đa thức đặt ẩn phụ :
a)
Đặt \(t=x^2-3x-1\)
\(t^2-12t+27\)
\(=t^2-3t-9t+27\)
\(=t\left(t-3\right)-9\left(t-3\right)\)
\(=\left(t-3\right)\left(t-9\right)\)
\(=\left(x^2-3x-1-3\right)\left(x^2-3x-1-9\right)\)
\(=\left(x^2-3x-4\right)\left(x^2-3x-10\right)\)
\(=\left(x^2+x-4x-4\right)\left(x^2-5x+2x-10\right)\)
\(=\left[x\left(x+1\right)-4\left(x+1\right)\right]\left[x\left(x-5\right)+2\left(x-5\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-5\right)\)
b,
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\)
\(t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=t^2-4t+6t-24\)
\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t-4\right)\left(t+6\right)\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a,\(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5x^2+5y^2-20x^2y^2\)
\(=5\left(x^2+y^2+x^2y^2\right)\)
Bài 2:
a,\(\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)(*)
Đặt \(x^2-3x-1=0\)
(*)\(=t^2-12t+27\)
\(=t^2-3t-9t+27\)
\(=t\left(t-3\right)-9\left(t-3\right)\)
\(=\left(t-9\right)\left(t-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
b,\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+x+4x+4\right)\left(x^2+2x+3x+6\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24\)
\(=\left(x^2+5x+5\right)^2-1^2-24\)
\(=\left(x^2+5x+5\right)^2-25\)
\(=\left(x^2+5x+5\right)^2-5^2\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
