Question

Phân tích đa thức thành nhân tử:
a) \(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc\)

b) \(abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\)

c) \(-6x^2-5y+3xy+10x\)

d) \(x^2+y^2-2xy-x+y\)

Answer 1

d, \(x^2+y^2-2xy-x+y\)

\(=\left(x-y\right)^2-\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-1\right)\)

Answer 2

c) \(-6x^2-5y+3xy+10x\)

\(=-\left(6x^2-3xy\right)+\left(10x-5y\right)\)

\(=-3x\left(2x-y\right)+5\left(2x-y\right)\)

\(=\left(2x-y\right)\left(-3x+5\right)\)

Answer 3

a) a(b² + c²) + b(a² + c²) + c(a² + b²) + 2abc
= ab² + ac² + bc² + ba² + a²c + b²c + abc + abc
= (ab² + b²c) + ( ac² + a²c) + ( bc² + abc) + ( ba² + abc)
= b² ( a + c) + ac ( a + c) + bc ( c + a) + ba ( a + c)
= ( a + c ) (b² + ac + bc + ba)
= ( a + c ) (( b²+ bc ) + ( ac + ba ))
= ( a + c ) ( b( b+c) + a ( c + a))
= ( a + c ) ( b+ c ) (b + a)

b) abc - ( ab + bc + ca ) + (a + b + c) -1
= abc - ab - bc - ca + a + b + c -1
= ( abc - ab ) - (bc - b) - (ac - a) + ( c -1)
= ab( c - 1) - b ( c - 1) - a( c - 1) + ( c -1)
= ( c -1)(ab - b - a +1)
= ( c -1)( (ab -b ) - ( a -1))
= ( c -1)( b( a -1) - ( a -1))
= ( c -1)( a -1)(b -1)

Answer 4

b) \(abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\)

\(=abc-ab-bc-ca+a+b+c-1\)

\(=abc-ab-bc+b-ca+a+c-1\)

\(=ab\left(c-1\right)-b\left(c-1\right)-a\left(c-1\right)+\left(c-1\right)\)

\(=\left(ab-b-a+1\right)\left(c-1\right)\)

\(=\left[b\left(a-1\right)-\left(a-1\right)\right]\left(c-1\right)\)

\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

Answer 5

c) \(-6x^2-5y+3xy+10x\)

\(=3xy-6x^2-5y+10x\)

\(=3x\left(y-2x\right)-5\left(y-2x\right)\)

\(=\left(3x-5\right)\left(y-2x\right)\)