Phân tích đa thức thành nhân tử:
a) \(\left(x-z\right)^2-y^2+2y-1\)
b) \(x^3+y^3+3y^2+3y+1\)
c) \(1-2a+2bc+a^2-b^2-c^2\)
d) \(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
e) \(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
Phân tích đa thức thành nhân tử: a) \(\left(x-z\right)^2-y^2+2y-1\) b) \(x^3+y^3+3y^2+3y+1\) c) \(1-2a+2bc+a^2-b^2-c^2\) d) \(x^2+3cd\left(2-3cd\right
By Gianna
Đáp án:
Câu e bn bảo ko lm nx nên mk ko lm
Giải thích các bước giải:
a) ( x-z)² -y² +2y -1= (x -z)² -(y-1)²
= ( x-z +y -1).( x-z-y+1)
c) 1 -2a +2bc +a² -b² -c²
= ( a² -2a+1) -( b² -2bc+c²)
= (a -1)² – ( b -c)²
= ( a-1+b+c).( a-1-b+c)
d) x² +3cd.( 2 -3cd) -10xy -1 +25y²
= ( x² -10xy +25y²) +( 6cd -9c²d² -1)
= ( x -5y)² – [(3cd)² -2.3.cd -1]
= ( x -5y)² – ( 3cd -1)²
= ( x -5y +3cd -1).( x-5y -3cd +1)
b) x³ +y³ +3y² +3y +1
= x³ +(y+1)³
= ( x +y+1).( x² – xy +x +y² +2y +1)
a, \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x-z\right)^2-\left(y-1\right)^2\)
\(=\left(x-z-y+1\right)\left(x-z+y-1\right)\)
b, \(x^3+y^3+3y^2+3y+1\)
\(=x^3+\left(y+1\right)^3=\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\)
\(=\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\)
c, \(1-2a+2bc+a^2-b^2-c^2\)
\(=a^2-2a+1-\left(b^2-2bc+c^2\right)\)
\(=\left(a-1\right)^2-\left(b-c\right)^2\)
\(=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
d, \(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)+\left[3cd\left(2-3cd\right)-1\right]\)
\(=\left(x-5y\right)^2+\left(6cd-\left(3cd\right)^2-1\right)\)
\(=\left(x-5y\right)^2-\left(3cd-1\right)^2\)
\(=\left(x-5y-3cd+1\right)\left(x-5y+3cd-1\right)\)
mình ko làm câu e theo ý của bn r ạ