Question

+ d) 2 radical din 2^11
h) radical din 2^5x3^4x5^7


scoateți factorii de sub radical

urgent!!!! ​

Answer 1

Răspuns:

Explicație pas cu pas:

a)

$\sqrt{2^3} = \sqrt{2^2\cdot 2} = 2\sqrt{2}$

b)

$\sqrt{2^5} = \sqrt{2^4\cdot 2} = 2^2\sqrt{2} = 4\sqrt{2}$

c)

$\sqrt{3^7} = \sqrt{3^6\cdot 3} = 3^3\sqrt{3} = 27\sqrt{3}$

d)

$2\sqrt{2^{11}} = 2\sqrt{2^{10}\cdot 2} = 2\cdot2^5\sqrt{2} = 2^6\sqrt{2} = 64\sqrt{2}$

e)

$-6\sqrt{2^3} =-6 \sqrt{2^2\cdot 2} = -6\cdot 2\sqrt{2} = -12\sqrt{2}$

f)

$\sqrt{2^3\cdot3^2} = \sqrt{2^2\cdot 2\cdot3^2} = 2\cdot3\sqrt{2} = 6\sqrt{2}$

g)

$-\sqrt{3^4\cdot5^7} = -\sqrt{3^4\cdot 5^6\cdot5} =- 3^2\cdot5^3\cdot\sqrt{5} =-9\cdot125 \sqrt{5}=-1125\sqrt{5}$

h)

$\sqrt{2^5\cdot3^4 \cdot5^7} = \sqrt{2^4\cdot 2\cdot3^4\cdot5^6\cdot5} = 2^2\cdot3^2\cdot5^3\cdot\sqrt{2\cdot5} = 4\cdot9\cdot125\sqrt{10} = 4500\sqrt{10}$

i)

$\sqrt{3^6\cdot5^5 \cdot11^2} = \sqrt{3^6\cdot 5^4\cdot5\cdot11^2} = 3^3\cdot5^2\cdot11\sqrt{5} = 27\cdot25\cdot11\sqrt{5} = 7425\sqrt{5}$

j)

$\sqrt{2^{2n+1}\cdot3^{4n} } = \sqrt{2^{2n}\cdot 2\cdot3^{4n}} = 2^n\cdot3^{2n}\cdot\sqrt{2}$

Answer 2

$\it d)\ 2\sqrt{2^{11}}=2\sqrt{2^{10}\cdot2}=2\cdot\sqrt{(2^5)^2\cdot2}=2\cdot2^5\sqrt2=2^6\sqrt2=64\sqrt2$

$\it h)\ \ \sqrt{2^5\cdot3^4\cdot5^7}=\sqrt{2\cdot2^4\cdot(3^2)^2\cdot5\cdot5^6}=\sqrt{2\cdot5\cdot(2^2)^2\cdot(3^2)^2\cdot(5^3)^2}=\\ \\ =2^2\cdot3^2\cdot5^3\sqrt{10}=4\cdot9\cdot125\sqrt{10}=4500\sqrt{10}$