Lời giải:

a) $\sqrt[3]{9}\cdot \sqrt[3]{27} = \sqrt[3]{3^2}\cdot \sqrt[3]{3^3} = 3^{2/3 + 1} = 3^{5/3} = 3\sqrt[3]{3^2}$

b) $\dfrac{\sqrt[4]{128}}{\sqrt{2}} = \dfrac{\sqrt[4]{2^7}}{2^{1/2}} = 2^{7/4 - 1/2} = 2^{5/4} = \sqrt[4]{32} = 4$

c) $\sqrt[3]{\sqrt[3]{9}} = 9^{1/9} = 3^{2/9} = \sqrt[9]{3^2} = \sqrt[9]{9}$

d) $\sqrt{2+\sqrt{162}-\sqrt{32}} = \sqrt{2+9\sqrt{2}-4\sqrt{2}} = \sqrt{2+5\sqrt{2}} = \sqrt{2}(\sqrt{1+\tfrac{5}{2}\sqrt{2}})$

e) $(\sqrt[3]{5})^4\cdot \sqrt[8]{81\sqrt[4]{5}} = 5^{4/3} \cdot (3^4\cdot 5^{1/4})^{1/8} = 5^{4/3+1/32}\cdot 3^{1/2} = 3^{1/2}\cdot 5^{43/32}$

Lời giải:

a) $\dfrac{3^{x-1}}{3^{x+1}} = 3^{(x-1)-(x+1)} = 3^{-2} = \dfrac{1}{9}$

b) $(4^x)^{3/4}\cdot \left(\dfrac{1}{4^x}\right)^{1/4} = 4^{3x/4}\cdot 4^{-x/4} = 4^{x/2} = 2^x$

c) $3^{2x+4}\cdot 3^{5-2x}\cdot 3^{-8} = 3^{2x+4+5-2x-8} = 3^1 = 3$

d) $\dfrac{(a^{4/5}b^{3/5})^5}{a^4b} = \dfrac{a^4b^3}{a^4b} = b^2$

Lời giải:

Ta có $8^x = (2^3)^x = 2^{3x}$ và $8^{-x} = 2^{-3x}$, do đó:

$\\dfrac{8^x - 8^{-x}}{2^x - 2^{-x}} = \\dfrac{2^{3x}-2^{-3x}}{2^x-2^{-x}} = 2^{2x} + 1 + 2^{-2x}$

Mà $4^x = 2^{2x} = 5$, nên $2^{-2x} = \\dfrac{1}{5}$.

Vậy giá trị biểu thức là $5 + 1 + \\dfrac{1}{5} = \\dfrac{31}{5}$.

Lời giải:

Ta có $5^x = 2 \\Rightarrow 5 = 2^{1/x}$ và $10^y = 2 \\Rightarrow 10 = 2^{1/y}$.

Suy ra: $2^{1/x} = 5$, $2^{1/y} = 10 = 2\cdot5 \\Rightarrow 2^{1/y} = 2^{1 + 1/x}$.

Do đó $\\dfrac{1}{y} = 1 + \\dfrac{1}{x} \\Rightarrow \\dfrac{1}{x} + \\dfrac{1}{y} = -1.$