9. (1)已知$2^m× 32× 4^m=2^{20}$,求$(-m^3)^2÷ (-m)^3$的值;
(2)若$10^m = 20$,$10^n=\dfrac{1}{5}$,求$9^n÷ 3^{2m}$的值;
(3)已知$x = 2^{m + 1}$,$y = 3 + 4^m$,用含$x$的代数式表示$y$.
(2)若$10^m = 20$,$10^n=\dfrac{1}{5}$,求$9^n÷ 3^{2m}$的值;
(3)已知$x = 2^{m + 1}$,$y = 3 + 4^m$,用含$x$的代数式表示$y$.
答案：9.(1)
∵2ᵐ×32×4ᵐ=2³ᵐ⁺⁵=2²⁰,
∴3m + 5 = 20,解得m = 5,
∴(−m³)²÷(−m)³=m⁶÷(−m)³=−m³=−5³=−125
(2)由10ᵐ=20,10ⁿ=$\frac{1}{5}$,得10ᵐ÷10ⁿ=20÷$\frac{1}{5}$=100,即10ᵐ⁻ⁿ=10²,
∴m - n = 2,
∴2m - 2n = 4,
∴2n - 2m = -4,
∴9ⁿ÷3²ᵐ=3²ⁿ÷3²ᵐ=3²ⁿ⁻²ᵐ=3⁻⁴=$\frac{1}{81}$
(3)由x = 2ᵐ⁺¹,y = 3 + 4ᵐ,得2ᵐ=$\frac{x}{2}$,y = 3 + (2ᵐ)²,
∴y = 3 + ($\frac{x}{2}$)²=$\frac{1}{4}$x² + 3
∵2ᵐ×32×4ᵐ=2³ᵐ⁺⁵=2²⁰,
∴3m + 5 = 20,解得m = 5,
∴(−m³)²÷(−m)³=m⁶÷(−m)³=−m³=−5³=−125
(2)由10ᵐ=20,10ⁿ=$\frac{1}{5}$,得10ᵐ÷10ⁿ=20÷$\frac{1}{5}$=100,即10ᵐ⁻ⁿ=10²,
∴m - n = 2,
∴2m - 2n = 4,
∴2n - 2m = -4,
∴9ⁿ÷3²ᵐ=3²ⁿ÷3²ᵐ=3²ⁿ⁻²ᵐ=3⁻⁴=$\frac{1}{81}$
(3)由x = 2ᵐ⁺¹,y = 3 + 4ᵐ,得2ᵐ=$\frac{x}{2}$,y = 3 + (2ᵐ)²,
∴y = 3 + ($\frac{x}{2}$)²=$\frac{1}{4}$x² + 3
10. 求等式中$x$的值:$3^{3x + 1}× 5^{3x + 1}=15^{2x + 4}$.
答案：10.
∵3³ˣ⁺¹×5³ˣ⁺¹=15³ˣ⁺¹,3³ˣ⁺¹×5³ˣ⁺¹=15²ˣ⁺⁴,
∴15³ˣ⁺¹=15²ˣ⁺⁴,
∴3x + 1 = 2x + 4,解得x = 3
∵3³ˣ⁺¹×5³ˣ⁺¹=15³ˣ⁺¹,3³ˣ⁺¹×5³ˣ⁺¹=15²ˣ⁺⁴,
∴15³ˣ⁺¹=15²ˣ⁺⁴,
∴3x + 1 = 2x + 4,解得x = 3
11. (1)已知$2^a× 27^b× 37^c× 47^d = 1998$,其中$a$,$b$,$c$,$d$是自然数,求$(a - b - c + d)^{203}$的值;
(2)试确定$3^{101}× 7^{102}× 13^{103}$的个位数字.
(2)试确定$3^{101}× 7^{102}× 13^{103}$的个位数字.
答案：11.(1)
∵2ᵃ×27ᵇ×37ᶜ×47ᵈ=2ᵃ×3³ᵇ×37ᶜ×47ᵈ,1998 = 2×3³×37,
∴2ᵃ×3³ᵇ×37ᶜ×47ᵈ=2×3³×37,
∴a = 1,b = 1,c = 1,d = 0,
∴(a - b - c + d)²⁰³=(1 - 1 - 1 + 0)²⁰³= -1
(2)3¹⁰¹×7¹⁰²×13¹⁰³=3¹⁰⁰×3×7¹⁰²×13¹⁰²×13=(3⁴)²⁵×3×(7×13)¹⁰²×13=81²⁵×91¹⁰²×39.
∵81²⁵,91¹⁰²的个位数字均为1,
∴81²⁵×91¹⁰²×39的个位数字为9,
∴3¹⁰¹×7¹⁰²×13¹⁰³的个位数字为9
∵2ᵃ×27ᵇ×37ᶜ×47ᵈ=2ᵃ×3³ᵇ×37ᶜ×47ᵈ,1998 = 2×3³×37,
∴2ᵃ×3³ᵇ×37ᶜ×47ᵈ=2×3³×37,
∴a = 1,b = 1,c = 1,d = 0,
∴(a - b - c + d)²⁰³=(1 - 1 - 1 + 0)²⁰³= -1
(2)3¹⁰¹×7¹⁰²×13¹⁰³=3¹⁰⁰×3×7¹⁰²×13¹⁰²×13=(3⁴)²⁵×3×(7×13)¹⁰²×13=81²⁵×91¹⁰²×39.
∵81²⁵,91¹⁰²的个位数字均为1,
∴81²⁵×91¹⁰²×39的个位数字为9,
∴3¹⁰¹×7¹⁰²×13¹⁰³的个位数字为9