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a. y4 - 14y2 + 49
Gọi y2 là t, ta có:
t2 - 14t + 49
<=> t2 - 14t + 72
<=> (t - 7)2
Thay x2 = t
<=> (x2 - 7)2
b. \(\dfrac{1}{4}-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c. x4 - 16
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
d. x2 - 9
<=> x2 - 32
<=> (x - 3)(x + 3)
a) ( x + 2 )( x2 - 2x + 4 ) - ( 18 + x3 )
= x3 + 8 - 18 - x3 = -10
b) ( 2x - y )( 4x2 + 2xy + y2 ) - ( 2x + y )( 4x2 - 2xy + y2 )
= 8x3 - y3 - ( 8x3 + y3 )
= 8x3 - y3 - 8x3 - y3 = -2y3
c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )
= x2 - 9 - ( x2 + 4x - 5 )
= x2 - 9 - x2 - 4x + 5 = -4x - 4
d) ( 3x - 2 )2 + ( x + 1 )2 + 2( 3x - 2 )( x + 1 )
= ( 3x - 2 + x + 1 )2
= ( 4x - 1 )2
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
a) (x+2)2+x(x-4)
=x2+4x+4+x2-4x
=2x2+4
b)(x-3)2-(x+3)(x-4)
=x2-6x+9-x2+4x-3x+12
=-5x+12
c) (3x+1)2+3x(2-4x)
=9x2+6x+1+6x-12x2
=-3x2+12x+1
d) (2x-4y)2-(2x-3)(2x-3y)
=4x2-16xy+16y2-4x2+6xy+6x-9y
=16y2-10xy+6x-9y
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)