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\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
Tham khảo nhé~
a , 3x2 + 3y2 - 6xy - 12
= 3 ( x2 + y2 - 2xy - 4 )
= 3 ( x - y )2 - 22
= 3 ( x - y + 2 ) ( x - y - 2 )
Bài giải:
a) x2 + 4x – y2 + 4 = (x2 + 4x + 4) - y2
= (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
b) 3x2 + 6xy + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y – z)(x + y + z)
c) x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
48. Phân tích các đa thức sau thành nhân tử:
a) x2 + 4x – y2 + 4; b) 3x2 + 6xy + 3y2 – 3z2;
c) x2 – 2xy + y2 – z2 + 2zt – t2.
Bài giải:
a) x2 + 4x – y2 + 4 = (x2 + 4x + 4) - y2
= (x + 2)2 – y2 = (x + 2 – y)(x + 2 + y)
b) 3x2 + 6xy + 3y2 – 3z2 = 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2] = 3(x + y – z)(x + y + z)
c) x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
a) \(x^2+4x-y^2+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
x4-3x3-x+3 = (x4-3x3)-(x-3) = x3(x-3)-(x-3) = (x-3)(x3-1) = (x-3)(x-1)(x2+x+1)
3x+3y-x2-2xy-y2 = (3x+3y)-(x2+2xy+y2) = 3(x+y)-(x+y)2 = (x+y)( 3-x-y)
x2-x-12 = x(x-1)-12
a) x2 +x -y2 + y = ( x2 -y2 ) +(x+y)
= (x-y)(x+y) +(x+y)
=(x+y)( x-y+1)
b) 3x2 +3y2 -6xy -12 = 3(x2 +y2 - 2xy) -12
=3 [ (x-y)2 -4]
= 3( x-y-2)(x-y+2)
a) x2 + x - y2 + y
= (x2 - y2) + (x + y)
= (x + y) (x - y) + (x + y)
= x + y
b) 3x2 + 3y2 - 6xy - 12
= 3 (x2 + y2 - 2xy - 4)
= 3 [(x2 - 2xy + y2) - 4]
= 3 [(x - y)2 - 22]
= 3 (x - y + 2) (x - y - 2)
(sai thì thôi)
a) x^3−3x^2−4x+12
=(x^3-3x^2)-(4x-12)
=x^2(x-3)-4(x-3)
=(x-3)(x^2-4)=(x-3)(x-2)(x+2)
b) x^4-5x^2+4=x^4-x^2-4x^2+4
=(x^4-x^2) - ( 4x^2-4)
=x^2(x^2-1) - 4(x^2-1)
=(x^2-1)(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
c) (x+y+z)^3-x^3-y^3-z^3
=x^3+y^3+z^3+3x^2yz+3xy^2z+3xyz^2-x^3-y^3-z^3
=3x^2yz+3xy^2z+3xyz^2
3xyz(x+y+z)
a) \(x^2+x-y^2+y\)
\(=\left(x-y\right)\left(x+y\right)+x+y\)
\(=\left(x+y\right)\left(x-y+1\right)\)
b) \(3x^2+6xy+3y^2-12\)
\(=3\left(x^2+2xy+y^2-4\right)\)
\(=3\left(x+y-2\right)\left(x+y+2\right)\)
c) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x+2\right)\left(x-2\right)\)
a)\(x^2+x-y^2+y\)
=\(\left(x^2-y^2\right)+\left(x+y\right)\)
=\(\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
=\(\left(x-y+1\right)\left(x+y\right)\)
b)\(3x^2+6xy+3y^2-12\)
=\(3\left(x^2+2xy+y^2-4\right)\)
=\(3\left[\left(x+y\right)^2-2^2\right]\)
=\(3\left(x+y-2\right)\left(x+y+2\right)\)
c)\(x^3-3x^2-4x+12\)
=\(x^2\left(x-3\right)-4\left(x-3\right)\)
=\(\left(x^2-4\right)\left(x-3\right)\)
=\(\left(x-2\right)\left(x+2\right)\left(x-3\right)\)