整式的除法~快来@!计算-3(ab)^2 × (-ab)^3 ÷ (12a^3b^2)2×2^n÷2^n-1+(-1又1/2)^8×(-2)^1016^2m ÷ 8^2n ÷ 4^m ×4^(n-m+1)(2mn)^2 × (m^2n^2)-(m^2n^2)^3 ÷m^3n^4+3m^2n^4(4x^n-1 y^n+2)^2÷(-x^n-2 y^n+1)(x^2a+3b+4c)^m ÷ [(x^a)^2m ×(x^3)^bm ×
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整式的除法~快来@!计算-3(ab)^2 × (-ab)^3 ÷ (12a^3b^2)2×2^n÷2^n-1+(-1又1/2)^8×(-2)^1016^2m ÷ 8^2n ÷ 4^m ×4^(n-m+1)(2mn)^2 × (m^2n^2)-(m^2n^2)^3 ÷m^3n^4+3m^2n^4(4x^n-1 y^n+2)^2÷(-x^n-2 y^n+1)(x^2a+3b+4c)^m ÷ [(x^a)^2m ×(x^3)^bm ×
整式的除法~快来@!
计算
-3(ab)^2 × (-ab)^3 ÷ (12a^3b^2)
2×2^n÷2^n-1+(-1又1/2)^8×(-2)^10
16^2m ÷ 8^2n ÷ 4^m ×4^(n-m+1)
(2mn)^2 × (m^2n^2)-(m^2n^2)^3 ÷m^3n^4+3m^2n^4
(4x^n-1 y^n+2)^2÷(-x^n-2 y^n+1)
(x^2a+3b+4c)^m ÷ [(x^a)^2m ×(x^3)^bm ×(x^m)^4c]
已知x^2-3x-2=0 求-x^3+11x+6
若a+b=0 ab=11 求a^2-ab+b^2
整式的除法~快来@!计算-3(ab)^2 × (-ab)^3 ÷ (12a^3b^2)2×2^n÷2^n-1+(-1又1/2)^8×(-2)^1016^2m ÷ 8^2n ÷ 4^m ×4^(n-m+1)(2mn)^2 × (m^2n^2)-(m^2n^2)^3 ÷m^3n^4+3m^2n^4(4x^n-1 y^n+2)^2÷(-x^n-2 y^n+1)(x^2a+3b+4c)^m ÷ [(x^a)^2m ×(x^3)^bm ×
-3(ab) ²× (-ab) ³ ÷ (12a³ b ²)
=3a²b²×a³b³÷12a³b²
=1/4•a²b³
2×2^n÷2^(n-1)+(-1又1/2)^8×(-2)^10
=2^[1+n-(n-1)+3^8/2 ^8×2^10
=2^2+3^8×2^2
=26248
16^2m ÷ 8^2n ÷ 4^m ×4^(n-m+1)
=2^8m ÷ 2^6n÷ 2^2m ×2^2(n-m+1)
=2^[8m-6n-2m+2(n-m+1)]
=2^(4m-4n+2)
(2mn) ²× (m²n²)-(m²n²)³÷m³ n^4+3m²n^4
=4m²n²× (m²n²)-(m^6 n^6)÷m³ n^4+3m²n^4
=4m^4 n^4-m^3 n^2+3m²n^4
[4x^(n-1) y^(n+2)] ²÷[-x^(n-2) y^(n+1)]
=16 x^2(n-1) y^2(n+2) ÷[-x^(n-2) y^(n+1)]
=-16 x^n y^(n+3)
[x^(2a+3b+4c)]^m ÷ [(x^a)^2m ×(x³)^bm ×(x^m)^4c]
= x^m(2a+3b+4c) ÷ [x^2ma ×x^3bm ×x^4c m]
= x^m(2a+3b+4c) ÷x^m(2a+3b+4c)
=1
已知x²-3x-2=0 求-x³+11x+6
因为x²-3x-2=0
故：x²=3x+2
故：-x³+11x+6
=-x(3x+2) +11x+6
=-3x²+9x+6
=-3(3x+2) +9x+6
=0
若a+b=0 ab=11 求a²-ab+b²
a²-ab+b²=(a+b) ²-3ab=-33
我很想回答，可惜我不会...
分少了！