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a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
Bài 1 :
a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)
b) \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)
c) \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
d) \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)
BÀi 2 :
a) \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)
\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)
b) \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)
c) \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)
\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)
\(=\left(b+c-a\right)\left(d-c^2\right)\)
BÀi 3 :
a) \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)
b) \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)
c) \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)
d) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
Ta có:
a) 6x2y - 3y2 - 2x2 + y = (6x2y - 2x2) - (3y2 - y) = 2x2(3y - 1) - y(3y - 1) = (2x2 - y)(3y - 1)
b) 2x2 + x - 4xy - 2y + 2x + 1 = (x2 + x) - (4xy + 2y) + (x2 + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)2
= (x + x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1 - 2y)
c) 16x2y - 4xy2 - 4x3 + x2y = 4xy(4x - y) - x2(4x - y) = (4xy - x2)(4x - y)
d) 4x2 - 20x + 25 - 36y2 = (2x - 5)2 - (6y)2 = (2x - 5 - 6y)(2x - 5 + 6y)
e) x2 - 4y2 + 6x - 4y + 8 = (x2 + 6x + 9) - (4y2 + 4y + 1) = (x + 3)2 - (2y + 1)2 = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)
g) Ta có : x10 + x5 + 1
= (x10 - x) + (x5 - x2) + (x2 + x + 1)
= x(x9 - 1) + x2(x3 - 1) + (x2 + x + 1)
= x(x3 - 1)(x6 + x3 + 1) + x2(x3 - 1) + (x2 + x + 1)
= (x7 + x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x8 - x7 + x 5 - x4 + x2 - x + x4 + x3 + x2 + 1)
= (x2 + x + 1)(x8 - x7 + x5 + x3 - x + 1)
h) TT trên (dài dòng ktl)
a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
C) x2+4x+4-9y2
= ( x+2 ) \(^2\) - 9y\(^2\)
= ( x+ 2 - 3y ) ( x + 2 + 3y)
D) x2-5x-6
= x\(^2\) + x - 6x - 6
= x( x + 1 ) - 6( x + 1 )
= ( x - 6 ) ( x+ 1 )