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a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
a) x3+y3+z3-3xyz
=(x+y)3+z3-3x2y-3xy2-3xyz
=(x+y+z).[(x+y)2+(x+y).z+z2]-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2)-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2-3xy)
=(x+y+z)(x2+y2+zx+zy+z2-zy)
b)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(-a2c+c2a)+(b2c-b2a)
=b.(a2-c2)-ac.(a-c)-b2.(a-c)
=b.(a+c)(a-c)-ac.(a-c)-b2.(a-c)
=(a-c)[b.(a+c)-ac-b2]
=(a-c)(ab+bc-ac-b2)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
a) \(\left(xy+1\right)^2-\left(x+y\right)\)
\(=\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
\(=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
a) ( xy + 1 )2 - ( x + y )2
= [ ( xy + 1 ) - ( x + y ) ][ ( xy + 1 ) + ( x + y ) ]
= ( xy - x - y + 1 )( xy + x + y + 1 )
b) ( x + y )3 - ( x - y )3
C1. = x3 + 3x2y + 3xy2 + y3 - ( x3 - 3x2y + 3xy2 - y3 )
= x3 + 3x2y + 3xy2 + y3 - x3 + 3x2y - 3xy2 + y3
= 6x2y + 2y3
= 2y( 3x2 + y2 )
C2. = [ ( x + y ) - ( x - y ) ][ ( x + y )2 + ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y - x + y )( x2 + 2xy + y2 + x2 - y2 + x2 - 2xy + y2 )
= 2y( 3x2 + y2 )
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= 3( x4y2 + x3y2 + xy2 + y2 )
= 3[ ( x4y2 + x3y2 ) + ( xy2 + y2 ) ]
= 3[ x3y2( x + 1 ) + y2( x + 1 ) ]
= 3( x + 1 )( x3y2 + y2 )
= 3y2( x + 1 )( x3 + 1 )
= 3y2( x + 1 )( x + 1 )( x2 - x + 1 )
= 3y2( x + 1 )2( x2 - x + 1 )
d) 4( x2 - y2 ) - 8( x - ay ) - 4( a2 - 1 )
= 4[ ( x2 - y2 ) - 2( x - ay ) - ( a2 - 1 )
= 4( x2 - y2 - 2x + 2ay - a2 + 1 )
= 4[ ( x2 - 2x + 1 ) - ( y2 - 2ay + a2 ) ]
= 4[ ( x - 1 )2 - ( y - a )2 ]
= 4[ ( x - 1 ) - ( y - a ) ][ ( x - 1 ) + ( y + a ) ]
= 4( x - y + a - 1 )( x + y + a - 1 )
1) (3x+4)(x+1) = 3x2+7x+4 đặt là a
(6x+7)2= 36x2+84x+49 = 12a+1
=> a(12a+1)- 6 = 12a2 -a -6 = (3a+2)(4a-3) = (9x2+21x+14)(12x2+28x+13)
2) (x-2)2=x2-4x+4 đặt là a
(2x-5)(2x-3)= 4x2-16x+15 =4a-1
=> a(4a-1)-5 = 4a2-a-5 = (4a-5)(a+1) = ( 4x2-16x+11)(x2-4x+5)
3) đặt (x+3)2 =a ta làm tương tự
4) (x-2)(x-10)(x-4)(x-5) = (x2-12x+20)(x2-9x+20)
đặt x2+20=a => (a-12x)(a-9x)-54x2 = a2-21ax+54x2 = (a-18x)(a-3x) = (x2-18x+20)(x2-3x+20)
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
a) Ta có: \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x-3\right)\left(x+y\right)\)
b) Ta có: \(x^3+10x^2+25x-xy^2\)
\(=x\left(x^2+10x+25-y^2\right)\)
\(=x\left(x+5-y\right)\left(x+5+y\right)\)
c) Ta có: \(x^3+2+3\left(x^3-2\right)\)
\(=4x^3-4\)
\(=4\left(x-1\right)\left(x^2+x+1\right)\)