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Phân tích đa thức sau thành nhân tử:
(x2+3x+1)(x2+3x+2)-6
Mình đang cần gấp, mong mọi người giải giùm.
1, x^2 +5x+4x+20
= x(x+4)+5(x+4)
= (x+5).(x+4)
2,x^2 -x -20
= x^2 -5x+4x-20
= x(x-5)+4(x-5)
=(x+4).(x-5)
3,3x^2 +3x -2x -2
= 3x(x+1)-2(x+1)
= (3x-2)(x+1)
4, 2x^2-4x-x-2
=2x(x-2)-x-2
=(2x-1)(x+2)
5,6x^2-2x+9x-3
=2x(3x-1)+3(3x-1)
=(2x+3)(3x-1)
6,3x^2 +2x+9x+6
=3x(x+3)+2(x+3)
=(3x+2)(x+3)
7,= 2(x^2-10x+3)
8,=x^4 +4^3
mk làm luôn ko chép đề nha bn
\(x^2+9x+20\)
\(\Leftrightarrow x^2+4x+5x+20\)
\(\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\)
\(x^2-x-20\)
\(\Leftrightarrow x^2-5x+4x-20\)
\(\Leftrightarrow\left(x^2-5x\right)+\left(4x-20\right)\)
\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)\)
\(3x^2+x-2\)
\(\Leftrightarrow3x^2+3x-2x-2\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(2x+2\right)\)
\(\Leftrightarrow3x\left(x+1\right)-2x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2x\right)\)
\(\)
\(\left(4-x\right)^2+\left(x-4\right)\left(x-5\right)-4\left(x-5\right)^2+1\)
= \(16-4x+x^2+x^2-5x-4x+20-4\left(x^2-5x+25\right)+1\)
= \(37-13x+2x^2-4x^2+20x+100\)
= \(137+7x-2x^2\)
\(=\left(x-4\right)^2+\left(x-4\right)\left(x-5\right)-\left(2\left(x-5\right)\right)^2+1\)
\(=\left(x-4\right)\left(2x-9\right)-\left(\left(2x-10\right)^2-1\right)\)
\(=\left(x-4\right)\left(2x-9\right)-\left(2x-11\right)\left(2x-9\right)\)
\(=\left(2x-9\right)\left(x-4-2x+11\right)=\left(2x-9\right)\left(7-x\right)\)
1) \(25x^4-10x^2y+y^2\)
\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)
\(\Leftrightarrow\left(5x^2+y\right)^2\)
2) \(x^4+2x^3-4x-4\)
\(\Leftrightarrow\left(x^4-4\right)+\left(2x^3-4x\right)\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^4+x^2+1\)
\(\Leftrightarrow x^4+x^2-x+x+1\)
\(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)
\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)
5) \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)
\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)
\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)
a) x2 - 16 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 16
= ( x - 2y )2 - 42
= ( x - 2y - 4 )( x - 2y + 4 )
b) x5 - x4 + x3 - x2
= x2( x3 - x2 + x - 1 )
= x2[ x2( x - 1 ) + ( x - 1 ) ]
= x2( x - 1 )( x2 + 1 )
c) x( x + 4 )( x + 6 )( x + 10 ) + 128 < mình nghĩ là nên sửa đề như này :]>
= [ x( x + 10 ) ][ ( x + 4 )( x + 6 ) ] + 128
= ( x2 + 10x )( x2 + 10x + 24 ) + 128
Đặt t = x2 + 10x
bthuc <=> t( t + 24 ) + 128
= t2 + 24t + 128
= t2 + 16t + 8t + 128
= t( t + 16 ) + 8( t + 16 )
= ( t + 16 )( t + 8 )
= ( x2 + 10x + 16 )( x2 + 10x + 8 )
= ( x2 + 2x + 8x + 16 )( x2 + 10x + 8 )
= [ x( x + 2 ) + 8( x + 2 ) ]( x2 + 10x + 8 )
= ( x + 2 )( x + 8 )( x2 + 10x + 8 )
cảm ơn bạn câu c mình chép nhầm nó là 128 đó
(x2 + x + 2)(x2 + 9x + 18) - 28
= x4 + 10x3 + 29x2 + 36x + 36 - 28
= x4 + 10x3 + 29x2 + 36x + 8
1) x2- 3x - 6x +18
= (x2- 3x )-(6x -18 )
= x(x-3)- 6(x-3)
= (x-6)(x-3)
\(2,x^2-3x-54\)
\(=x^2-9x+6x-54\)
\(=x\left(x-9\right)+6\left(x-9\right)\)
\(=\left(x+6\right)\left(x-9\right)\)
\(3,20x^2+7x-6\)
\(=20x^2-8x+15x-6\)
\(=4x\left(5x-2\right)+3\left(5x-2\right)\)
\(=\left(4x+3\right)\left(5x-2\right)\)