Rozwiązane

(x²+2)²-(x-2)(x+2)(x²+4)=

(a²-2)²-(a+1)(a-1)(a²+1)=

(2y-1)²(y-2)-(2y-3)(2y+3)y+3y²=

(2y²-3b)²-(2y-b)(2y+b)y+12by²+4y³=



Odpowiedź :

(x²+2)²-(x-2)(x+2)(x²+4)=x⁴+4x²+4-(x²-4)(x²+4)=x⁴+4x²+4-x⁴+16=4x²+20

(a²-2)²-(a+1)(a-1)(a²+1)= a⁴-4a²+4-(a²-1)(a²+1)= a⁴-4a²+4-a⁴+1= -4a²+5

(2y-1)²(y-2)-(2y-3)(2y+3)y+3y²=
(4y²-4y+1)(y-2)-(4y²-9)y+3y²=
4y³-4y²+y-8y²+8y-2-4y³+9y+3y²=
-9y+18y-2

(2y²-3b)²-(2y-b)(2y+b)y+12by²+4y³=
4y⁴-12by²+9b² - (4y²-b²)y+12by²+4y³=
4y⁴-12by²+9b²-4y³+b²y+12by²+4y³=
4y⁴+b²y+9b²
XklaudysiaX
1) = 8x²²²² (to jest 8x do potęgi 8)
2) = 5a³³ (to jest 5a do potęgi 6)
3) = -48y
4) = 41y×18b

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