Реши уравнение 2r−3r2−121−r−3r2−11r=r−1r2+11r

Question

Алгебра: Реши уравнение 2r−3r2−121−r−3r2−11r=r−1r2+11r

boha3 · Accepted Answer

1)12a²b³+3a³b²+16b²-a²=3a²b²(4b+a) + (4²b² - a²)=3a²b²(4b+a)+(4b-a)(4b+a)=

= (4b+a)(3a²b² + 4b- a)

2) 49c² -14c+1 -21ac+3a = (49c²-14c+1) -3a(7c - 1) = (7c - 1)² - 3a(7c - 1) =

=(7c-1)(7c - 1 - 3a)

3)ax²+ay²+x^4+2x²y²+y^4 = a(x²+y²)+(x^4+2x²y²+y^4) = a(x²+y²) +(x²+y²)²=

= (x²+y²) (a +x²+y²)

4) 27c³-d³+9c²+3cd+d² = [(3c)³-d³]+ (9c²+3cd+d²) =

=[(3c - d)(9c²+3cd+d²)] + (9c²+3cd+d²) = (9c²+3cd+d²) (3c-d+1)

5) b³-2b²-2b+1 =(b³ + 1) - 2b( b+1) = (b+1)(b² -b+1) - 2b(b+1) =

= (b+1)(b² -b+1-2b) = (b+1)(b² -3b+1)

MOGZ ответил

Реши уравнение
2r−3r2−121−r−3r2−11r=r−1r2+11r