Phân tích đa thức thành phân tử
\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
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(x^2-10xy+25y^2)-(1-3cd(2-3cd))
=(x^2-2.x.5.y+(5y)2)-(1-6cd+9(cd)2)
=(x-5)^2-((3cd)2-2.3cd.1+12)
=(x-5)^2-(3cd-1)2
=(x-5+3cd-1).(x-5-3cd+1)
=(x+3cd-6).(x-3cd-4)...
Theo mk là z. Có thể phân tích nữa,đúng hay sai mk cx chưa chắc chắn vì mk cũng mới học thui.
a)(x-z)^2-y^2+2y-1=(x-z)^2-(y^2-2y+1) b)1-2a+2bc+a^2-b^2-c^2=a^2-2a+1-b^2+2bc-c^2
=(x-z)^2-(y-1)^2 =(a-1)^2-(b^2-2bc+c^2)
=(x-z-y+1)(x-z+y-1) =(a-1)^2-(b-c)^2
=(a-1-b+c)(a-1+b-c)
c)x^2+y^2-2xy-x+y=x^2-2xy+y^2-(x-y)
=(x-y)^2-(x-y)
=(x-y)(x-y-1)
1)
b) \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x^2-2xz+z^2\right)-\left(y-1\right)^2\)
\(=\left(y-z\right)^2-\left(y-1\right)^2\)
\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)
\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)
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, \(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)+\left[3cd\left(2-3cd\right)-1\right]\)
\(=\left(x-5y\right)^2+\left(6cd-\left(3cd\right)^2-1\right)\)
\(=\left(x-5y\right)^2-\left(3cd-1\right)^2\)
\(=\left(x-5y-3cd+1\right)\left(x-5y+3cd-1\right)\)
Chúc bạn học tốt!!!
Giúp bạn câu e nè
e) bc ( b + c) + ac ( c - a) - ab ( a + b)
= b2 c + bc2 + ac 2 - a2c - a2b - ab2
= ( b2 c - a2c ) + ( bc2+ ac2 ) - ( a2b + ab2 )
= c ( b2- a2 ) + c2 ( b+ a ) - ab(a + b)
= c( b - a )( b + a) + c2 ( b+ a ) - ab(a + b)
= ( b+ a )( bc - ac + c2- ab )
= ( b+ a )( ( bc - ab ) - ( ac - c2 ))
= ( b+ a )( b( c - a ) - c ( a - c))
= ( b+ a )( b( c - a ) + c ( c - a))
= ( b+ a )( c - a )(b + c )
\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2=x^2+6cd-\left(3cd\right)^2-10xy-1+\left(5y\right)^2\\ \)
\(=x^2-10xy+\left(5y\right)^2-\left(1-6cd+\left(3cd\right)^2\right)\)
\(=\left(x-5y\right)^2+6cd-1-\left(3cd\right)^2=\left(x-5y\right)^2-\left(1-3cd\right)^2\)
\(=\left(x-5y-1+3cd\right)\left(x-5y+1-3cd\right)\)