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Câu 1:
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(0.5x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(0.5x-y-2x-2y\right)\left(0.5x-y+2x+2y\right)\)
\(=\left(-1.5y-3y\right)\left(2.5x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(0.2x+0.2y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x+2y+\dfrac{1}{5}y+\dfrac{1}{5}x\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x+\dfrac{11}{5}y\right)\)
d: \(=-\left(5x-3\right)^2\)
a: \(=\left(5x+10y\right)^2-\left(8x-4y\right)^2\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(-3x+14y\right)\left(13x+6y\right)\)
b: \(=\left(\dfrac{1}{2}x-y\right)^2-\left(2x+2y\right)^2\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
c: \(=\left(\dfrac{2}{3}x-2y\right)^2-\left(\dfrac{1}{5}x+\dfrac{1}{5}y\right)^2\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
a)x4-1=(x2-1)(x2+1)=(x-1)(x+1)(x2+1)
b)x2-y2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)
c)x2-6x-y2+9=(x2-6x+9)-y2=(x-3)2-y2=(x-y-3)(x+y-3)
d)5x2+3(x+y)2-5y2
=5(x2-y2)+3(x+y)2
=5(x-y)(x+y)+3(x+y)2
=(x+y)(5x-5y+3x+3y)
=(x+y)(8x-2y)
Áp dụng HĐT a2 - b2 = ( a - b )( a + b )
và tính chất an.bn = ( a.b )n ( với n ∈ N* )
a) ( 3x + 1 )2 - ( x + 1 )2
= [ ( 3x + 1 ) - ( x + 1 ) ][ ( 3x + 1 ) + ( x + 1 ) ]
= ( 3x + 1 - x - 1 )( 3x + 1 + x + 1 )
= 2x( 4x + 2 )
= 2x.2( 2x + 1 )
= 4x( 2x + 1 )
b) ( x + y )2 - ( x - y )2
= [ ( x + y ) - ( x - y ) ][ ( x + y ) + ( x - y ) ]
= ( x + y - x + y )( x + y + x - y )
= 2y.2x = 4xy
c) ( 2xy + 1 )2 - ( 2x + y )2
= [ ( 2xy + 1 ) - ( 2x + y ) ][ ( 2xy + 1 ) + ( 2x + y ) ]
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )9 y + 1 )( 2x + 1 )
d) 9( x - y )2 - 4( x + y )2
= 32( x - y )2 - 22( x + y )2
= [ 3( x - y ) ]2 - [ 2( x + y ) ]2
= ( 3x - 3y )2 - ( 2x + 2y )2
= [ ( 3x - 3y ) - ( 2x + 2y ) ][ ( 3x - 3y ) + ( 2x + 2y ) ]
= ( 3x - 3y - 2x - 2y )( 3x - 3y + 2x + 2y )
= ( x - 5y )( 5x - y )
e) ( 3x - 2y )2 - ( 2x - 3y )2
= [ ( 3x - 2y ) - ( 2x - 3y ) ][ ( 3x - 2y ) + ( 2x - 3y ) ]
= ( 3x - 2y - 2x + 3y )( 3x - 2y + 2x - 3y )
= ( x + y )( 5x - 5y )
= ( x + y )5( x - y )
f) ( 4x2 - 4x + 1 ) - ( x + 1 )2
= ( 2x - 1 )2 - ( x + 1 )2
= [ ( 2x - 1 ) - ( x + 1 ) ][ ( 2x - 1 ) + ( x + 1 ) ]
= ( 2x - 1 - x - 1 )( 2x - 1 + x + 1 )
= 3x( x - 2 )
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
a) 5x-15y=5x-3.5.y=5(x-3y)
c) 14xy(xy+28x)
d) \(\dfrac{2}{7}\left(3x-1\right)\left(x-1\right)\)
e) (x-1)3
f) (x+y-2x)(x+y+2x)=(y-x)(3x+y)
g) (3x+\(\dfrac{1}{2}\))(9x2+\(\dfrac{3}{2}x\)+\(\dfrac{1}{4}\))
h) (x+y-x+y)\(\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
2a)
(x+1)(x2+2x)=0
(x+1)x(x+2)=0
\(\left[{}\begin{matrix}x+1=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)
a) x2 - 2x - 4y2 - 4y
= (x2 - 4y2) - (2x + 4y)
= (x + 2y)(x - 2y) - 2(x + 2y)
= (x + 2y)(x - 2y - 2)
= (x + 2y)[x - 2(y + 1)]
b) x4 + 2x3 - 4x - 4
= (x4 - 4) + ( 2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
c) x3 + 2x2y - x -2y
= (x3 - x) + (2x2y - 2y)
= x(x2 - 1) + 2y(x2 - 1)
= (x + 2y)(x2 - 1)
a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
Bài 1:
a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)
\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)
\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(14y-3x\right)\left(13x+6y\right)\)
b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)
c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)
\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)
d) \(-25x^2+30x-9\)
\(=-\left(25x^2-30x+9\right)\)
\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)
\(=-\left(5x-3\right)^2\)
Bài 2:
a) \(x^3y^2-x^2y^3-2x+2y\)
\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2y^2-2\right)\)
Thay x = -1 và y = -2 vào ta được
\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)
\(=1\left(4-2\right)\)
\(=2\)
b) \(5x^2-3x+3y-5y^2\)
\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
Thay x = 3 và y = 1 vào ta được
\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)
\(=5.2.4-3.2\)
\(=34\)