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20. (4分)已知4ᵐ = 5,8ⁿ = 3,3ᵐ = 4,计算下列代数式的值：
(1)2²ᵐ⁺³ⁿ; (2)2⁴ᵐ⁻⁶ⁿ; (3)12²ᵐ.
解:(1)2²ᵐ⁺³ⁿ = 2²ᵐ·2³ⁿ = 5×3 = 15.
(2)2⁴ᵐ⁻⁶ⁿ = 2⁴ᵐ÷2⁶ⁿ = (2²ᵐ)²÷(2³ⁿ)² = .
(3)12²ᵐ = (3×4)²ᵐ = 3²ᵐ×4²ᵐ = (3ᵐ)²×(4ᵐ)² = 4²×5² = 16×25 = 400.
(1)2²ᵐ⁺³ⁿ; (2)2⁴ᵐ⁻⁶ⁿ; (3)12²ᵐ.
解:(1)2²ᵐ⁺³ⁿ = 2²ᵐ·2³ⁿ = 5×3 = 15.
(2)2⁴ᵐ⁻⁶ⁿ = 2⁴ᵐ÷2⁶ⁿ = (2²ᵐ)²÷(2³ⁿ)² = .
(3)12²ᵐ = (3×4)²ᵐ = 3²ᵐ×4²ᵐ = (3ᵐ)²×(4ᵐ)² = 4²×5² = 16×25 = 400.
答案：
$4^{m}=2^{2m}=5,8^{n}=2^{3n}=3,3^{m}=4$.
(1)$2^{2m + 3n}=2^{2m}\cdot2^{3n}=5×3 = 15$.
(2)$2^{4m - 6n}=2^{4m}\div2^{6n}=(2^{2m})^{2}\div(2^{3n})^{2}=\frac{25}{9}$.
(3)$12^{2m}=(3×4)^{2m}=3^{2m}×4^{2m}=(3^{m})^{2}×(4^{m})^{2}=4^{2}×5^{2}=16×25 = 400$.
(1)$2^{2m + 3n}=2^{2m}\cdot2^{3n}=5×3 = 15$.
(2)$2^{4m - 6n}=2^{4m}\div2^{6n}=(2^{2m})^{2}\div(2^{3n})^{2}=\frac{25}{9}$.
(3)$12^{2m}=(3×4)^{2m}=3^{2m}×4^{2m}=(3^{m})^{2}×(4^{m})^{2}=4^{2}×5^{2}=16×25 = 400$.
21. (4分)(1)已知xᵃ = 2,xᵇ = 5,求xᵃ⁺ᵇ的值;
解:∵xᵃ = 2,xᵇ = 5,∴xᵃ⁺ᵇ = xᵃ·xᵇ = 2×5 = 10.
(2)已知3²·9²ˣ⁺¹÷27ˣ⁺¹ = 81,求x的值.
解:∵3²·9²ˣ⁺¹÷27ˣ⁺¹ = 3²·3⁴ˣ⁺²÷3³ˣ⁺³
= 3²⁺⁴ˣ⁺²⁻(³ˣ⁺³) = 3ˣ⁺¹ = 81 = 3⁴,
∴x + 1 = 4,解得x = 3.
解:∵xᵃ = 2,xᵇ = 5,∴xᵃ⁺ᵇ = xᵃ·xᵇ = 2×5 = 10.
(2)已知3²·9²ˣ⁺¹÷27ˣ⁺¹ = 81,求x的值.
解:∵3²·9²ˣ⁺¹÷27ˣ⁺¹ = 3²·3⁴ˣ⁺²÷3³ˣ⁺³
= 3²⁺⁴ˣ⁺²⁻(³ˣ⁺³) = 3ˣ⁺¹ = 81 = 3⁴,
∴x + 1 = 4,解得x = 3.
答案：
(1)$\because x^{a}=2,x^{b}=5,\therefore x^{a + b}=x^{a}\cdot x^{b}=2×5 = 10$.
(2)$\because 3^{2}\cdot9^{2x + 1}\div27^{x + 1}=3^{2}\cdot3^{4x + 2}\div3^{3x + 3}=3^{2 + 4x + 2-(3x + 3)}=3^{x + 1}=81 = 3^{4}$,$\therefore x + 1 = 4$,解得$x = 3$.
(1)$\because x^{a}=2,x^{b}=5,\therefore x^{a + b}=x^{a}\cdot x^{b}=2×5 = 10$.
(2)$\because 3^{2}\cdot9^{2x + 1}\div27^{x + 1}=3^{2}\cdot3^{4x + 2}\div3^{3x + 3}=3^{2 + 4x + 2-(3x + 3)}=3^{x + 1}=81 = 3^{4}$,$\therefore x + 1 = 4$,解得$x = 3$.
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