Question

Помогите пожалуйста , 35 баллов даю !

Answer 1

Ответ:

Объяснение:

1) ...=pq(2p*(3p-q)-(6q²-2pq)) = pq(6q²-2pq-6q²+2pq) = pq(6q²-6q²) = pq*6(p²-q²) = pq*6(p-q)(p+q) = 6pq(p-q)(p+q)

2)... = mn(3m(mn-2)-(3mn²-6m)) = mn(3m²n-6m-(3mn²-6m)) = mn(3m²n-6m-3mn²+6m) = mn(3m²n-3mn²) = mn*3mn(m-n) = 3m²n²(m-n)

3) ... = 2b(a²b²-b²a+a³)-ab²(ab-2b+a²) = b(2(a²b²-ab²+a³)-a²b²+2ab²-a³b) = b(2a²b²-2ab²+2a³-a²b²+2ab²-a³b) = b(a²b²-2a³-a³b) = a²b(b²+2a-ab)

1) ... = 0,5x²yz*2x³-0,5x²yz*3y³+0,5x²yz*z² = x⁵yz-1,5x²y³z+0,5x²yz³

2) ... = 1/3mn²p*3m²-1/3mn²p*6n³-1/3mn²p*p² = mn²pm²-mn²p*2n³-1/3mn²p³ = m³n²p-2mn⁵p-1/3mn²p³

/ - дробь

Answer 2

Ответ:

Объяснение:

1) 2p²q(3p-q)-pq(6q²-2pq)=2p²q(3p-q)--2pq²(3q-p)=(3q-p)(2p²q-2pq²)=2pq(3q-p)(p-q)=2pq(3pq-3q²-p²+pq)=2pq(-p²+4pq-3q²)=-2p³q+8p²q²-6pq³

2) 3m²n(mn-2)-mn(3mn²-6m)=3m²n(mn-2)-3m²n(n²-2)=3m²n(mn-2-n²+2)=3m²n(mn-n²)=3m²n²(m-n)=3m³n²-3m²n³

3) (a²b²-b²a+a³)·2b-(ab-2b+a²)·ab²=2ab(a²+ab²-b²)-ab²(a²+ab-2b)=ab(2(a²+ab²-b²)-b(a²+ab-2b))=ab(2a²+2ab²-2b²-a²b-ab²+2b²)=ab(2a²-a²b+ab²)=a²b(2a-ab+b²)=2a³b-a³b²+a²b³=-a³b²+2a³b+a²b³

4) (xy²-y³-xy)·3x²-xy(2x²y-3xy²-1)=3x²y(xy-y²-x)-xy(2x²y-3xy²-1)=xy(3x(-x+xy-y²)-2x²y+3xy²+1)=xy(-3x²+3x²y-3xy²-2x²y+3xy²+1)=xy(-3x²+x²y+1)=-3x³y+x³y²+xy=x³y²-3x³y+xy

1) 0,5x²yz(2x³-3y²+z²)=(x²yz(2x³-3y²+z²))/2=(2x⁵yz-3x²y³z+x²yz³)/2

2) 1/3 ·mn²p(3m²-6n³-p²)=(mn²p(3m²-6n³-p²))/3=(3m³n²p-6mn⁵p-mn²p³)/3