1) ac2-ad+c3-cd-bc2+bd= = (ac2 – ad) + (c3 – bc2) + (bd – cd) = a·(c2 – d) + c2·(c – b) + d·(b – c) = a·(c2 – d) + c2·(c – b) – d·(c – b) = a·(c2 – d) + c2·(c – b) – d·(c – b) = a·(c2 – d) + (c – b)·(c2 – d) = (c2 – d)·(a + c – b)
2) mx2+my2-nx2-ny2+n-m= x2 ( m - n ) + y2 ( m - n ) - ( m - n ) = ( m-n ) (x2 + y2 - 1 )
3) am2+cm2-an+an2-cn+cn2= m2 (a + c ) + n2 ( a + c ) - n ( a + c ) = ( a+ c) ( m2 + n2 - n)
4) xy2-ny2-mx+mn+m2x-m2n= y2 ( x - n ) + m2 ( x - n) - m ( x - n ) = ( x-n) ( y2 + m2 - m )
5) a2b+a+ab2+b+2ab+2=ab ( a + b + 2 ) + ( a+ b+ 2 ) = 2 ( a+ b + 2 )
6) x2-xy+x-xy2+y3-y2= x ( x – y + 1) – y 2 ( x – y + 1)=( x – y + 1)( x – y 2 ).
(х - 1)(х + 2) = 10
х² - х + 2х - 2 = 10
х² + х - 2 - 10 = 0
х² + х - 12 = 0
D = b² - 4ac = 1² - 4 · 1 · (-12) = 1 + 48 = 49
√D = √49 = ±7
х = (-b±√D)/2a
х₁ = (-1-7)/(2·1) = (-8)/2 = -4
х₂ = (-1+7)/(2·1) = 6/2 = 3
ответ: б) 3 - корень уравнения.