Question

BARDZO PROSZE O POMOC​

Answer 1

Verified answer

Odpowiedź:

a)

(9¹⁾³ * 27¹⁾²)/√3 = [(3²)¹⁾³ * (3³)¹⁾²]/3¹⁾² = (3²⁾³ * 3³⁾²)/3¹⁾² = 3²⁾³⁺³⁾²/3¹⁾² =

= 3⁴⁾⁶⁺⁹⁾⁶/3¹⁾² = 3¹³⁾⁶/3¹⁾² = 3¹³⁾⁶⁻¹⁾² = 3¹³⁾⁶⁻³⁾⁶ = 3¹⁰⁾⁶ = 3⁵⁾³ = 3³⁾³ * 3²⁾³ =

= 3¹ * 3²⁾³ = 3 * ∛3² = 3∛9

b)

⁴√4³ * ⁴√9 * ⁴√6⁶ = ⁴√(4³ * 9 * 6⁶) = ⁴√(64 * 3² * 3⁶ * 2⁶) =

= ⁴√(2⁶ * 2⁶ + 3² * 3⁶) = ⁴√(2⁶⁺⁶ * 3²⁺⁶) = ⁴√(2¹² * 3⁸) = ⁴√2¹² * ⁴√3⁸ =

= ⁴√(2⁴)³ * ⁴√(3⁴)² = 2³ * 3² = 8 * 9 = 72

Answer 2

[tex]a)\\\\\dfrac{9^{\frac{1}{3}}\cdot27^{\frac{1}{2}}}{\sqrt{3}}=\dfrac{(3^2)^{\frac{1}{3}}\cdot(3^3)^{\frac{1}{2}}}{3^{\frac{1}{2}}}=\dfrac{3^{\frac{2}{3}}\cdot3^{\frac{3}{2}}}{3^{\frac{1}{2}}}=\dfrac{3^{\frac{2}{3}+\frac{3}{2}}}{3^{\frac{1}{2}}}=\dfrac{3^{\frac{4}{6}+\frac{9}{6}}}{3^{\frac{1}{2}}}=\dfrac{3^{\frac{13}{6}}}{3^{\frac{1}{2}}}=3^{\frac{13}{6}-\frac{1}{2}}=\\\\\\=3^{\frac{13}{6}-\frac{3}{6}}=3^{\frac{10}{6}}=3^{\frac{5}{3}}[/tex]

[tex]b)\\\\\sqrt[4]{4^3}\cdot\sqrt[4]{9}\cdot\sqrt[4]{6^6}=\sqrt[4]{4^3\cdot9\cdot6^6}=\sqrt[4]{(2^2)^3\cdot3^2\cdot3^6\cdot2^6}=\sqrt[4]{2^6\cdot2^6\cdot3^2\cdot3^6}=\\\\=\sqrt[4]{2^1^2\cdot3^8}=\sqrt[4]{2^1^2}\cdot\sqrt[4]{3^8}=\sqrt[4]{(2^4)^3}\cdot\sqrt[4]{(3^4)^2}=2^3\cdot3^2=8\cdot9=72[/tex]

[tex]Zastosowane\ \ wzory\\\\a^m\cdot a^n=a^{m+n}\\\\\frac{a^m}{a^n}=a^{m-n}\\\\(a^m)^n=a^{m\cdot n}\\\\\sqrt[n]{a}=a^{\frac{1}{n}}[/tex]