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a) (-x2 +6x3 - 26x + 21) : (3-2x)
= -3x2 + 5x + 11/2 ( dư 37/1/2)
b) (2x4 - 13x3 - 15 + 5x + 21x2) : (4x-x2 -3)
= -2x2 + 5x + 5
a) => 4x2y2 - (4x2.2) yz + 4x2z2
=> 4x2.(y2+yz+z2 - 2)
chắc sai!! 45454655474675675685685787686845765756856876
a) \(4x^2y^2-8x^2yz+4x^2z^2\)
\(=\left(2xy\right)^2-2.2xy.2xz+\left(2xz\right)^2\)
\(=\left(2xy-2xz\right)^2\)
\(=4x^2\left(y-z\right)^2\)
b) \(x^8+x^7+x^6+x^5+x^3\)
\(=x^3\left(x^5+x^4+x^3+x^2+1\right)\)( có lẽ vậy )
\(x^2-\left(y-3\right)^2-4x+4\)
\(=x^2-\left(y^2-6y+9\right)-4x+4\)
\(=x^2-y^2+6y-9-4x+4\)
\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)
\(=\left(x-2\right)^2-\left(y-3\right)^2\)
\(=\left[\left(x-2\right)-\left(y-3\right)\right]\left[\left(x-2\right)+\left(y-3\right)\right]\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
1.
x2 - ( y - 3 )2 - 4x + 4
= ( x2 - 4x + 4 ) - ( y - 3 )2
= ( x - 2 )2 - ( y - 3 )2
= [ ( x - 2 ) - ( y - 3 ) ][ ( x - 2 ) + ( y - 3 ) ]
= ( x - 2 - y + 3 )( x - 2 + y - 3 )
= ( x - y + 1 )( x + y - 5 )
2.
a) Ta có : 2x4 + 8x3 + 9x2 - 4x - 5
= 2x4 + 10x2 - x2 + 8x3 - 4x - 5
= ( 2x4 - x2 ) + ( 8x3 - 4x ) + ( 10x2 - 5 )
= x2( 2x2 - 1 ) + 4x( 2x2 - 1 ) + 5( 2x2 - 1 )
= ( 2x2 - 1 )( x2 + 4x + 5 )
=>(2x4 + 8x3 + 9x2 - 4x - 5) : ( 2x2 - 1 ) = x2 + 4x + 5
b) Ta có : x2 + 4x + 5 = ( x2 + 4x + 4 ) + 1 = ( x + 2 )2 + 1 ≥ 1 > 0 ∀ x
=> đpcm
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
1. 4-32x3
= 4.(1-8x3)
= 4.[13-(2x)3 ]
= 4.(1-2x).(1+2x+4x2)
2. b. \(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right):\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left[\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(x-y\right)}\right]:\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left[\frac{x.x}{y\left(x-y\right).x}+\frac{\left(2x-y\right).y}{x\left(x-y\right).y}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{x^2+2xy-y^2}{xy\left(x-y\right)}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{-\left(x-y\right)^2}{xy\left(x-y\right)}\right].\frac{xy}{x+y}\)
\(=\frac{-\left(x-y\right)}{xy}.\frac{xy}{x+y}\)
\(=\frac{y-x}{x+y}\)
phần 1 đề nhầm ak sửu lại nha:
\(\left(8x^3+1\right):\left(4x^2-2x+1\right)=\left(2x+1\right)\left(4x^2-2x+1\right):\left(4x^2-2x+1\right)=2x+1\)
2) \(x^2-y^2-6x+6y\)
\(=\left(x-y\right)\left(x+y\right)-6\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-6\right)\)
\(a,4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
a) Kết quả - 2 x 4 + 3 x 2 + 5. b) Kết quả - 4 x 3 + 1 2 x − 1.