www.psdd.net > 已知集合A=〔m丨㎡%3m+2=0〕,B〔m丨㎡%Am+4=...

已知集合A=〔m丨㎡%3m+2=0〕,B〔m丨㎡%Am+4=...

解: (m-1)(m-2)=0 m=1或m=2 A={1,2} B⊆A,则B可能为Φ,{1},{2},{1,2} B=Φ时,方程m²-am+4=0的判别式△

(1)am?an=am+n;(2)(-b)3÷(-b)=(-b)3-1=(-b)2=b2;(3)(-2a)3=(-2)3?a3=-8a3;(4)(a2)4=a2×4=a8;(5)(3m+2n)(3m-2n)=(3m)2-(2n)2=9m2-4n2;(6)(4m-2n)2=(4m)2-2?4m?2n+(2n)2=16m2-16mn+4n2.故答案为:a...

a^n=8 a^mn=64 (a^n)^m=8^2 8^m=8^2 m=2 a^m=3,a^n=5 a^(m-n) =a^m÷a^n =5/3 a^(3m-2n) =(a^3m)÷a^2n =(a^m)^3÷(a^n)^2 =3^3÷5^2 =27/25 2^x=1/32 2^x=(1/2)^5 2^x=2^(-5) x=-5 (2)若(-2)^x = (-2)^3÷(-2)^2x 则x= (-2)^x = (-2)^(3-2x) x=3-2...

(1)∵3×9m×27m=3×(32)m×(33)m=3×32m×33m=31+2m+3m∴1+2m+3m=16,∴m=3;(2)∵a3m+2n-k=a3m?a2n÷ak=(am)3?(an)2÷akam=2,am=2,an=4,ak=32,∴原式=23×42÷32=4.

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