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a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
a.\(4xy-8x^3y=4xy\left(1-2x^2\right)\)
b.\(5\left(x^2-y^2\right)+16\left(x-y\right)=\left(x-y\right)\left[5\left(x+y\right)+16\right]\)
c.\(x^3-4x^2y+4xy^3=x\left(x-2y\right)^2\)
d.\(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
e.\(x^2y-2x^2+4y-8=\left(y-2\right)\left(x^2+4\right)\)
g. \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
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)
a, 5x2 - 45x = 5x(x - 9)
b, 3x3y - 6x2y - 3xy3 - 6axy2 - 3a2xy + 3xy
= 3xy(x2 - 2x - y2 - 2ay - a2 + 1)
= 3xy[ (x2 - 2x + 1) - (a2 + 2ay + y2) ]
= 3xy[ (x - 1)2 - (a + y)2 ]
= 3xy(x - 1 + a + y)(x - 1 - a - y)
f, 3xy2 - 12xy + 12x
= 3x(y2 - 4y + 4)
= 3x(y - 2)2
g, 2x2 - 8x + 8
= 2(x2 - 4x + 4)
= 2(x - 2)2
h, 5x3 + 10x2y + 5xy2
= 5x( x2 + 2xy + y2 )
= 5x(x + y)2
k, x2 + 4x - 2xy - 4y + y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
i, x3 + ax2 - 4a - 4x
= (x3 - 4x) + (ax2 - 4a)
= x(x2 - 4) + a(x2 - 4)
= (x + a)(x2 - 4)
= (x + a)(x + 2)(x - 2)
Chúc bạn học tốt !
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) \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\) \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)
c) 2x^3y - 2xy^3 - 4xy^2 - 2xy
= 2xy ( x^2 - y^2 - 2y - 1 )
= 2xy ( x^2 - ( y^2 + 2y + 1 )
= 2xy ( x^2 - ( y + 1 )^2 )
= 2x ( x - y - 1 )( x + y + 1 )
sai bạn ơi !
đáp án là
= 2xy (x + y + 1) (x - y + 1)
that pun cho ban Nguyen Dieu Thao :((
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) xy – 3x + 2y – 6
= (xy - 3x) + (2y - 6)
= x(y - 3) + 2(y - 3)
= (y - 3)(x + 2)
b) x2y + 4xy + 4y – y3
= y(x2 + 4x + 4 - y2)
= y[(x2 + 4x + 4) - y2]
= y[(x + 2)2 - y2]
= y(x + 2 + y)(x + 2 - y)
c) x2 + y2 + xz + yz + 2xy
= (x2 + 2xy + y2) + (xz + yz)
= (x + y)2 + z(x + y)
= (x + y)(x + y + z)
d) x3 + 3x2 – 3x – 1
= (x3 - 1) + (3x2 - 3x)
= (x - 1)(x2 + x + z) + 3x(x - 1)
= (x - 1)(x2 + 4x + 1)
a )
\(xy-3x+2y-6\)
\(=\left(xy+2y\right)-3x-6\)
\(=y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(y-3\right)\left(x+2\right)\)
b )
\(x^2y+4xy+4y-y^3\)
\(=y\left(x^2+4x+4-y^2\right)\)
\(=y\left[\left(x+2\right)^2-y^2\right]\)
\(=y\left(x+2-y\right)\left(x+2+y\right)\)
c )
\(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
a. x2 + 4xy + 4y2 - 2x - 4y
= (x + 2y)2 - (2x + 4y)
= (x + 2y)2 - 2(x + 2y)
= (x + 2y)(x + 2y - 2)
= (x + 2y)[x + 2(y-1)]
b. x2 - 8x + 7
= x2 - x - 7x + 7
= x(x - 1) - (7x - 7)
= x(x - 1) - 7(x - 1)
= (x - 7)(x - 1)
c. xy - xz - y + z
= x(y - z) - (y - z)
= (x - 1)(y - z)
d. x3 + 4x2 + 4x
= x3 + 4x2 + 4x + x2 - x2
= x2(x + 1) + 4x(x + 1) - x2
= (x2 + 4x - x2)(x + 1)
= 4x(x + 1)