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a/ 5x2-x+5xy-y
= 5x(x+y) - (x+y)
= (5x-1) (x+y)
b) 2xz + 3z + 6y + xz
= ko bt làm
a; x(5x-1)+y(5x-1)
=>( x+y).(5x-1)
b; có ghi sai đề ko vậy ; toàn x vs z thế
a, =x4(x+2)-x3(x+2)+x2(x+2)-x(x+2)+(x+2)
=(x+2)(x4-x3+x2-x+1)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
a, 5x^2 +5xy - x - y
= 5x ( x+ y ) - (x + y)
= ( 5x - 1)(x + y)
b, 2x^2 + 3x - 5
= 2x^2 - 2x + 5x - 5
= 2x( x - 1) + 5( x - 1)
= ( 2x + 5 )(x- 1 )
c; 16x - 5x^2 - 3
c, = - ( 5x^2 - 16x + 3 )
= - ( 5x^2 - x - 15x + 3 )
= - [ x(5x - 1 ) - 3 (5x - 1) ]
= - ( x- 3)(5x - 1 )
- 9y3x – 36y2x= 9xy(y2–4) =9xy(y–4)(y+4)
- 64-y^2 – x^2 – 2xy = 64– (x^2 + 2xy + y^2) = 82 – (x+y)2 = (8 – x –y)(8+x+y)
- x^2 + x – 30= x^2 + x – 25 – 5 = (x2 – 25)(x – 5)= (x-5)(x+5)(x-5)= (x-5)^2 (x+5)
a)x6-y6+
=[(x2)3-(y2)3]+(x4+x2y2+y4)
=[(x2-y2)(x4+x2y2+y4)]+(x4+x2y2+y4)
=(x4+x2y2+y4)[(x2-y2)+1]
=(x2-xy+y2)(x2+xy+y2)(x2-y2+1)
a) 5x2 - 5xy - 3x + 3y
= 5x.(x - y) - 3.(x - y)
= (x - y).(5x - 3)
b) x3 - 2x2 - x + 2
= x2.(x - 2) - (x - 2)
= (x - 2).(x2 - 1)
= (x - 2).(x - 1).(x + 1)
a/ 5x2-5xy-3x+3y
=5x.(x-y)-3(x-y)
=(5x-3).(x-y)
b/ x3-2x2-x+2
=x2.(x-2)-(x-2)
=(x2-1).(x-2)
=(x-1).(x+1).(x-2)
a, x2-5xy+2x-10y = (x2 + 2x)-(5xy+10y)
= x(x+2)-5y(x+2)
= (x+2)(x-5y)
b, x2-5x+4 = x2- x - 4x +4
= (x2-x)-(4x-4)
=x(x-1)-4(x-4)
=(x-1)(x-4)
\(a,x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(x+2\right)\)
\(b,x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=x\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
Câu a đề sai
b
\(Q=y^4+64\)
\(Q=\left(y^4+16y^2+64\right)-16y^2\)
\(Q=\left(y^2+8\right)^2-\left(4y\right)^2\)
\(Q=\left(y^2+8-4y\right)\left(y^2+8+4y\right)\)