完全平方差(2x+y)(2x-y)+(x+2y)(x-2y)2004²-2003*2005(1+1/2)(1+

jacyhung2022-10-04 11:39:541条回答

完全平方差
(2x+y)(2x-y)+(x+2y)(x-2y)
2004²-2003*2005
(1+1/2)(1+1/2平方)(1+1/2四次方)(1+1/2八次方)+1/2十五次方
(5+1)(5²+1)(5四次方+1)````````(5六十四次方+1)
100²-99²+98²-97²+96²-95²-------+2²-1²
若x²-y²=6 x+y=2 求2x-y
118²-18²=
503²-497²=

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蓝莹雪梦3 共回答了15个问题 | 采纳率86.7%
(2x+y)(2x-y)+(x+2y)(x-2y)
=(2x)²-y²+x²-(2y)²
=4x²-y²+x²-4y²
=5x²-5y²
=5(x²-y²)
=5(x-y)(x+y)
2004²-2003×2005
=2004²-(2004-1)(2004+1)
=2004²-(2004²-1)
=2004²-2004²+1
=1
(1+1/2)(1+1/2²)(1+1/2^4)(1+1/2^8)+1/2^15
=2×(1-1/2)(1+1/2)(1+1/2²)(1+1/2^4)(1+1/2^8)+1/2^15
=2×(1-1/2²)(1+1/2²)(1+1/2^4)(1+1/2^8)+1/2^15
=2×(1-1/2^4)(1+1/2^4)(1+1/2^8)+1/2^15
=2×(1-1/2^8)(1+1/2^8)+1/2^15
=2×(1-1/2^16)+1/2^15
=2×1-2×1/2^16+1/2^15
=2-1/2^15+1/2^15
=2
(5+1)(5²+1)(5^4+1)……(5^64+1)
=1/4×(5-1)(5+1)(5²+1)(5^4+1)……(5^64+1)
=1/4×(5²-1)(5²+1)(5^4+1)……(5^64+1)
=1/4×(5^4-1)(5^4+1)……(5^64+1)
=1/4×(5^8-1)……(5^64+1)
=1/4×(5^64-1)(5^64+1)
=1/4×(5^128-1)
100²-99²+98²-97²+96²-95²+……+2²-1²
=(100+99)(100-99)+(98+97)(98-97)+(96+95)(96-95)+……+(2+1)(2-1)
=100+99+98+97+96+95+……+2+1
=(100+1)×100÷2
=5050
若x²-y²=6,x+y=2,则:
x-y=(x-y)(x+y)/(x+y)
=(x²-y²)/(x+y)
=6/2
=3
联立x+y=2,x-y=3 解方程组,得 x=5/2,y=-1/2
2x-y=2×5/2-(-1/2)
=5+1/2
=11/2
118²-18²
=(118-18)(118+18)
=100×136
=13600
503²-497²
=(503-497)(503+497)
=6×1000
=6000
1年前

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