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a) 2x.(4x - 1)
câu b), c) mik ko biết
ko mong b cho mik
nhưng vẫn hi vọng b hoặc ai đó sẽ làm vậy
b) \(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
a) \(8x^2-2x-1=8x^2-4x+2x-1=4x.\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(4x+1\right)\)
b) \(4x^4+1=\left(2x^2\right)^2+4x^2+1-4x^2=\left(2x^2+1\right)^2-4x^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
c) \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6=x^4-2x^3-x^2-2x^3+4x^2+2x-6\)
\(=x^4-4x^3+3x^2+2x-6=\left(x^4-3x^3\right)-\left(x^3-3x^2\right)+\left(2x-6\right)\)
\(=x^3.\left(x-3\right)-x^2.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x^3-x^2+2\right)\)
\(=\left(x-3\right)\left[\left(x^3+x^2\right)+\left(-2x^2-2x\right)+\left(2x+2\right)\right]\)
\(=\left(x-3\right)\left[x^2\left(x+1\right)-2x.\left(x+1\right)+2.\left(x+1\right)\right]=\left(x-3\right)\left(x+1\right)\left(x^2-2x+2\right)\)
a, 8x^2-2x-1 = 8x2-4x+2x-1 = 4x ( 2x -1) + (2x-1) = (4x+1)(2x-1)
b) 4x4+1 = (2x2)2 + 4x2+ 1 - 4x2 = (2x2+1)2-(2x)2 = (2x2+1-2x)(2x2+1+2x)
A = 6x4 - 5x3 + 4x2 + 2x - 1
= 6x4 + 3x3 - 8x3 - 4x2 + 8x2 + 4x - 2x - 1
= 3x3. ( 2x + 1 ) - 4x2 ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )
= ( 2x + 1 ) ( 3x3 - 4x2 + 4x - 1 )
= ( 2x + 1 ) ( 3x3 - x2 - 3x2 + x + 3x - 1 )
= ( 2x + 1 ) [ x2 ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]
= ( 2x + 1 ) ( 3x - 1 ) ( x2 - x + 1 )
B = 4x4 + 4x3 + 5x2 + 8x - 6
= 4x4 - 2x3 + 6x3 - 3x2 + 8x2 - 4x + 12x - 6
= 2x3 ( 2x - 1 ) + 3x2 ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )
= ( 2x - 1 ) ( 2x3 + 3x2 + 4x + 6 )
= ( 2x - 1 ) [ x2 ( 2x + 3 ) + 2 ( 2x + 3 ) ]
= ( 2x - 1 ) ( 2x + 3 ) ( x2 + 2 )
C = x4 + x3 - 5x2 + x - 6
= x4 - 2x3 + 3x3 - 6x2 + x2 - 2x + 3x - 6
= x3 ( x - 2 ) + 3x2 ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )
= ( x - 2 ) ( x3 + 3x2 + x + 3 )
= ( x - 2 ) [ x2 ( x + 3 ) + ( x + 3 ) ]
= ( x - 2 ) ( x + 3 ) ( x2 + 1 )
a 4x -4y +(x-y)^2
=4(x-y)+(x-y).(x-y)
=(x-y).(4+x-y)
c x^2(x+1)-4(x+1)
(x+1).(x^2-4)
d x^4-(x^2-2x+1)
=x^4-(x-1)^2
=x^2(x-x+1)(x-x-1)
MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM
Câu b dễ thôi
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
a) 2x.( 4x - 1)
b),c) mik ko biết
Ko mong bạn tk cho mik
nhưng mik vẫn hi vọng b sẽ làm vậy