Question

Jika a=6 dan b=-1, tentukan nilai dari bentuk aljabar berikut.
a. a²+2ab+b²
b. 2a+2a²b²+3ab²+b³
c. a⁴+4a³b+6a²b²+b³

Answer 1

A. a²+ 2ab+ b² = 36+(-12)+1 = 25
b. 2a+ 2a²b²+3ab²+b³ = 12+72+18+(-1) = 101
c. 1296+(-864)+216+(-1) = 647

Sorry, ku salah ngitung

Answer 2

A.
a²+2ab+b²
= 6²+2(6)(-1)+(-1)²
= 36 - 12 + 1
= 24+1
= 25

b.
2a+2a²b²+3ab²+b³
= 2(6)+2(6²)(-1)²+3(6)(-1)²+(-1)³
= 12 + 2(36)(1) + 18 + (-1)
= 12 + 72 + 17
= 101

c.
a⁴+4a³b+6a²b²+b³
= 6⁴+4(6)³(-1)+6(6)²(-1)²+(-1)³
= 1296 + 4(216)(-1) + 6(36)(1) + (-1)
= 1296 - 864 + 216 - 1
= 432 + 215
= 647

Maaf, kalau ada yang salah.

Answer 3

A.
a²+2ab+b²
= 6²+2(6)(-1)+(-1)²
= 36 - 12 + 1
= 24+1
= 25

b.
2a+2a²b²+3ab²+b³
= 2(6)+2(6²)(-1)²+3(6)(-1)²+(-1)³
= 12 + 2(36)(1) + 18 + (-1)
= 12 + 72 + 17
= 101

c.
a⁴+4a³b+6a²b²+b³
= 6⁴+4(6)³(-1)+6(6)²(-1)²+(-1)³
= 1296 + 4(216)(-1) + 6(36)(1) + (-1)
= 1296 - 864 + 216 - 1
= 432 + 215
= 647

Maaf, kalau ada yang salah.