Phân tích các đa thức sau thành nhân tử:
1, \(5x+5y-x^2+y^2\)
2, \(3x^2-12x+12\)
3, \(x^4-1\)
4, \(x^5-1\)
5, \(75-30a+3a^2-3b^2\)
6,\(ax^2-ay^2+bx+by\)
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\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
a: \(\left(3x-1\right)^2-16\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x+3\right)\left(3x-5\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
b: \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-2x-4\right)\left(12x-4\right)\)
\(=-8\left(x+2\right)\left(3x-1\right)\)
1) x2 - x - y2 - y = (x - y)(x + y) - (x + y) = (x - y - 1)(x + y)
2. x2 - 2xy + y2 - z2 = (x - y)2 - z2 = (x - y - z)(x - y + z)
3. 5x - 5y + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
4. a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
5. 4x2 - y2 + 4x + 1 = (2x + 1)2 - y2 = (2x + 1 - y)(2x + y + 1)
6. x3 - x + y3 - y = (x + y)(x2 - xy + y2) - (x + y) = (x + y)(x2 - xy + y2 - 1)
Trả lời:
1, x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
2, x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - x2
= ( x - y - z ) ( x - y + z )
3, 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x ( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
= ( 5 + a ) ( x - y )
4, a3 - a2x - ay + xy
= ( a3 - a2x ) - ( ay - xy )
= a2 ( a - x ) - y ( a - x )
= ( a - x ) ( a2 - y )
5, 4x2 - y2 + 4x + 1
= ( 4x2 + 4x + 1 ) - y2
= ( 2x + 1 )2 - y2
= ( 2x + 1 - y ) ( 2x + 1 + y )
6, x3 - x + y3 - y
= ( x3 + y3 ) - ( x + y )
= ( x + y ) ( x2 - xy + y ) - ( x + y )
= ( x + y ) ( x2 - xy + y - 1 )
\(ax+bx+ay+by\)
\(=x\left(a+b\right)+y\left(a+b\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(xy+1-x-y\)
\(=x\left(y-1\right)-\left(y-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)
\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
\(2)3x+4-x^2-4x\)
\(=3(x+4)-\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(3)5x^2-2y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)[5(x+y)-10]\)
Còn lại bn lm nốt nha!
\(a,3x^2+2x=x\left(3x+2\right)\)
\(b,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)
\(c,4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(d,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
\(e,x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
a ) 3x2 + 2x
= x. ( 3x + 2 )
b ) 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x.( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
c ) 4x2 - 25
= ( 2x + 5 ) ( 2x - 5 )
d ) x2 + 6x + 5
= x2 + x + 5x + 5
= x.( x + 1 ) + 5.( x + 1 )
= ( x + 1 ) ( x + 5 )
e ) x2 - y2 + 2y - 1
= x2 - ( y - 1 )2
= ( x - y + 1 ) ( x + y - 1 )
f ) x3 - 3x + 2
= x3 + 2x2 - 2x2 - 4x + x + 2
= x2 ( x + 2 ) - 2x ( x + 2 ) + ( x + 2 )
= ( x + 2 ) ( x2 - 2x + 1 )
= ( x + 2 ) ( x - 1 )2
1) \(=5\left(x+y\right)-\left(x-y\right)\left(x+y\right)=\left(x+y\right)\left(5-x+y\right)\)
2) \(=3\left(x^2-4x+4\right)=3\left(x-2\right)^2\)
3) \(=\left(x^2-1\right)\left(x^2+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
4) \(=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\)
5) \(=3\left(a^2-10a+25-b^2\right)=3\left(\left(a-5\right)^2-b^2\right)=3\left(a-5-b\right)\left(a-5+b\right)\)
6) \(=a\left(x-y\right)\left(x+y\right)+b\left(x+y\right)=\left(x+y\right)\left(ax-ay+b\right)\)