Вычислите ординату точку пересечения графиков функций у=2x-7 и y=3x+4

Question

Алгебра: Вычислите ординату точку пересечения графиков функций у=2x-7 и y=3x+4

doge232 · Accepted Answer

1) x³+x²y-xy²-y³=(x³-y³)+(x²y-xy²)=(x-y)(x²+xy+y²)+xy(x-y)=
=(x-y)(x²+xy+y²+xy)=(x-y)(x²+2xy+y²)=(x-y)(x+y)²

2) a²+2ab+b²-ac-bc=(a²+2ab+b²)-(ac+bc)=(a+b)²-c(a+b)=
=(a+b)(a+b-c)

3) m²+2mn+n²-p²+2pq-q²=(m²+2mn+n²)-(p²-2pq+q²)=
=(m+n)² - (p-q)² = (m+n-p+q)(m+n+p-q)

4) x⁵-x³+x²-1=(x⁵-x³)+(x²-1)=x³(x²-1)+(x²-1)=(x²-1)(x³+1)=
=(x-1)(x+1)(x+1)(x²-x+1)=(x+1)²(x-1)(x²-x+1)

5) a³-8+6a²-12a=(a³-8)+(6a²-12a)=(a³-2³)+6a(a-2)=
=(a-2)(a²+2a+4)+6a(a-2)=(a-2)(a²+2a+4+6a)=
=(a-2)(a²+8a+4)

6) a⁴+a³+a+1=(a⁴+a³)+(a+1)=a³(a+1)+(a+1)=(a+1)(a³+1)=
=(a+1)(a³+1³)=(a+1)(a+1)(a²-a+1)=(a+1)²(a²-a+1)

7) x³-x²y-xy²+y³=(x³+y³)-(x²y+xy²)=(x+y)(x²-xy+y²)-xy(x+y)=
=(x+y)(x²-xy+y²-xy)=(x+y)(x²-2xy+y²)=(x+y)(x-y)²

MOGZ ответил