先化简再求值:(1)(2a+b)² +(a-2b)²-2(a-2b)(2a+b),其中27²=a^6 =9^b

问题描述:

先化简再求值:(1)(2a+b)² +(a-2b)²-2(a-2b)(2a+b),其中27²=a^6 =9^b

27²=a^6
∴(3^3)^2=a^6
∴3^6=a^6
a=±3
27²=9^b
∴(3^)^2=(3^2)^3=9^3=9^b
∴b=3
(2a+b)² +(a-2b)²-2(a-2b)(2a+b)
=[(2a+b)-(a-2b)]²
=[2a+b-a+2b]²
=[a+3b]²
当a=3,b=3时,原式=(a+3b)²=(3+3×3)²=12²=144
当a=-3,b=3时,原式=(a+3b)²=(-3+3×3)²=6²=36

其中27²=a^6 =9^b
=>3^6=a^6=(3^b)²=27²
=>a=3,3^b=27=3^3
=>b=3
(1)(2a+b)² +(a-2b)²-2(a-2b)(2a+b)
=(2a+b-a+2b)²
=(a+3b)²
=(3+9)²
=144

∵27²=a^6
(3³)²=a^6
3^6=a^6
∴a=3
∵27²=9^b
(3³)²=9^b
(3²)³=9^b
9³=9^b
∴b=3
(2a+b)²+(a-2b)²-2(a-2b)(2a+b)
=[(2a+b)-(a-2b)]²
=(a+3b)²
=(3+9)²
=12²
=144
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27²=a^6 =9^b
27²=3^6
∴a=3
27²=9³
∴b=3

(2a+b)² +(a-2b)²-2(a-2b)(2a+b)
=[(2a+b)-(a-2b)]²
=(2a+b-a+2b)²
=(a+3b)²
=(3+9)²
=144