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Bài làm
a) xy + y2 - x - y
= ( xy + y2 ) - ( x + y )
= y( x + y ) - ( x + y )
= ( x + y )( y - 1 )
b) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 25 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
c) xy + xz - 2y - 2z
= ( xy + xz ) - ( 2y + 2z )
= x( y + z ) - 2( y + z )
= ( y + z )( x - 2 )
d) x2 - 6xy + 9y2 - 25z2
= ( x2 - 6xy + 9y2 ) - 25z2
= ( x - 3y )2 - 25z2
= ( x - 3y - 5z )( z - 3y + 5z )
e) 3x2 - 3y2 - 12x + 12y
= 3( x - y )( x + y ) - 12( x - y )
= ( x - y )[ 3( x + y ) - 12 ]
f) 4x3 + 4xy2 + 8x2y - 16x
= 4x( x2 + y2 + 2xy - 4 )
= 4x[ ( x + y)2 - 4 ]
= 4x( x + y - 2 )( x + y + 2 )
g) x2 - 5x + 4
= x2 - x - 4x + 4
= x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( x - 4 )
h) x4 + 5x2 + 4
= x4 + x2 + 4x2 + 4
= x2( x2 + 1 ) + 4( x2 + 1 )
= ( x2 + 1 )( x2 + 4 )
i) 2x2 + 3x - 5
= 2x2 - 5x + 2x - 5
= 2x( x + 1 ) - 5( x + 1 )
= ( x + 1 )( 2x - 5 )
k) x3 - 2x2 + 6x - 5 ( không biết làm )
l) x2 - 4x + 3
= ( x2 - 4x + 4 ) - 1
= ( x - 2 )2 - 1
= ( x - 3 )( x - 1 )
# Học tốt #
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
Bài 1:
b: =x^2-10x+x-10
=(x-10)(x+1)
c: \(=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\)
d: \(=3x^2+5x-3x-5=\left(3x+5\right)\left(x-1\right)\)
e: \(=\left(2x+y\right)^3\)
a) -4x2 + 8x - 4
= - (4x2 - 8x + 4)
= - (2x - 2)2
b) -x52 + 10 x - 5
= - 5(x2 - 2x + 1)
= - 5(x - 1)2
bài 1
a, \(x^2+9y^2-6xy=\left(x-3y\right)^2\)
thay x = 19 , y = 3 vào biểu thức trên ta có
\(\left(19-3.3\right)^2=100\)
b, \(x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
thay x = 12 và y = -4 vào biểu thức trên ta có
\(\left(12-2.\left(-4\right)\right)^3=8000\)
bài 4
a, \(x\left(4x^2-1\right)=0\)
=> \(x\left(2x-1\right)\left(2x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b, \(x^3-x^2-x+1=0\)
=> \(x^2\left(x-1\right)-\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c, \(2x^2-5x-7=0\)
=> \(2x^2-7x+2x-7=0\)
=> \(2x\left(x+1\right)-7\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(2x-7\right)=0\)
=> \(\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)
Bài 2: Rút gọn biểu thức:
a) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2+4xy+2y^2-x^2+y^2\)
\(=2y^2-2xy\)
b)\(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2-3\left(1+4x\right)\left(1-4x\right)\)
\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)
\(=8x^2+40x+50-3+48x^2\)
\(=56x^2+40x+47\)
a: \(x^3+xy^2-y^2-1\)
\(=\left(x^3-1\right)+y^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+y^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+y^2\right)\)
b: \(12x^2+4x-6xy-2y\)
\(=4x\left(3x+1\right)-2y\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x-2y\right)=2\left(2x-y\right)\left(3x+1\right)\)