Phân tích đa thức thành nhân
a,9(x-15)-16(x+y)
b,9x^2-12xy+4y^2
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\(9x^2-12xy+4y^2\)
\(=\left(3x\right)^2-2.3x.2y+\left(2y\right)^2\)
\(=\left(3x-2y\right)^2\)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)
`a, 9x^2 - 16 = (3x+4)(3x-4)`
`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`
`c, t^3-8 = (t-2)(t^2 - 2t + 4)`
`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`
a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)
d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)
Bài 1:
\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)
Bài 2:
\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)
a: \(ab+a+b+1\)
\(=a\left(b+1\right)+\left(b+1\right)\)
\(=\left(b+1\right)\left(a+1\right)\)
c: \(4x^2-12xy+3x-9y\)
\(=4x\left(x-3y\right)+3\left(x-3y\right)\)
\(=\left(x-3y\right)\left(4x+3\right)\)
phân tích đa thức thành nhân tử
a, 6x^2 + 7xy + 2y^2
=6x^2+3xy+4xy+2y^2
=3x(x+y)+2y(x+y)
=(3x+2y)(x+y)
b, 9x^2 - 9xy - 4y^2
=9x^2 +3xy-12xy-4y^2
=3x(x+y)-4y(x+y)
=(3x+4y)(x+y)
c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)
Bài làm
a, 6x2 + 7xy + 2y2
= 6x2 + 3xy + 4xy + 2y2
= ( 6x2 + 3xy ) + ( 4xy + 2y2 )
= 3x( 2x + y ) + 2y( 2x + y )
= ( 2x + y )( 3x + 2y )
b, 9x2 - 9xy - 4y2
= 9x2 - 12xy + 3xy - 4y2
= ( 9x2 - 12xy ) + ( 3xy - 4y2 )
= 3x( 3x - 4y ) + y ( 3x - 4y )
= ( 3x + y )( 3x - 4y )
c, x2 - y2 + 10x - 6y + 16
= x2 - y2 - 6x + 6y + 4x + 16
= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )
= ( x - y + 4 )( x + 6 )
# Học tốt #
b) 9x^2-12xy+4y^2
\(=\left(3x\right)^2-2.3x.2y+\left(2y\right)^2=\left(3x-2y\right)^2..\)