Question

Ajutor, nu știu cum sa îl rezolv!!!!!!!!!!!!
Apropo ambele ex​

Answer 1

Explicație:

• Ex. 5: Mezii sunt egali cu extremii (Adica diagonalele se inmultesc intre ele)

a) $\frac{x}{16}=\frac{4}{x}$ => $x*x = 16 *4$ => $x^{2} = 64 => x=\sqrt{64}=>x=8$

b) $\frac{4}{x} =\frac{x}{9} => x^{2} =36=>x=6$

c) $\frac{x}{5} = \frac{3}{x} => x^{2} = 15 => x=3,87$

d) $\frac{2,5}{x}=\frac{x}{8} => x^{2} =20=>x=4,47$

e) $x^{2} = 2\sqrt{225}=>x^{2} =30 =>x=5,47$

f)$\frac{x}{2\frac{3}{4} }= \frac{11}{x} =>\frac{x}{ 2,75} =\frac{11}{x} => x^{2} =30.25=> x=5.5$

• Ex. 6: Media geometrica al numerelor x si y este egala cu $\sqrt{x*y}$

a) $\sqrt{12*3}=\sqrt{36}=6$

b) $\sqrt{16} =$ 7.74

c) $\sqrt{144}=12$

d) $\sqrt{3\frac{1}{5}*5}=\sqrt{\frac{16}{5}*5}=\sqrt{16} = 7.74$

e) $\sqrt{\frac{7}{12}*2\frac{1}{3}}=\sqrt{\frac{7}{12}*\frac{7}{3}}=\sqrt{0.58*2,(3)} = \sqrt{3,48} =1,86$

f) $\sqrt{\sqrt{5}^{-1}*\sqrt{\frac{5}{16}}} = \sqrt{\sqrt{5^{-1}*\frac{5}{16}}}=\sqrt[4]{5^{-1}*\frac{5}{16}}=\sqrt[4]{\frac{1}{16}}=5$

Sper ca te-am ajutat!