Question

Ajutati ma la 6 și 7 va rog!Este urgent!!​

Answer 1

Răspuns:

deocamdata 6 si ii de la 7, ma mai gandesc la i de la 7

Explicație pas cu pas:

6.

$81< 96\\81<49*2\\3^4 < 7^2*2\\\sqrt[4]{3^4} < \sqrt[4]{7^2*2} \\3 < \sqrt[4]{7^2}*\sqrt[4]{2} \\3 < \sqrt{7}*\sqrt[4]{2}\\\frac{3}{\sqrt[4]{2}} < \sqrt{7}\\(\frac{3}{\sqrt[4]{2}})^2 < (\sqrt{7})^2\\(\frac{3}{\sqrt[4]{2}})^2 < 7\\log_3[(\frac{3}{\sqrt[4]{2}})^2] < log_3(7) \\2*log_3(\frac{3}{\sqrt[4]{2}}) < log_3(7) \\2*[log_3(3) - log_3(\sqrt[4]{2})] < log_3(7) \\log_3(9)*[1 - log_3(\sqrt[4]{2})] < log_3(7) \\log_3[9^{[1 - log_3(\sqrt[4]{2})}] < log_3(7) \\9^{[1 - log_3(\sqrt[4]{2})]}< 7$

7

ii

$(a-b)^2 \geq 0\\a^2 + b^2 - 2ab \geq 0\\a^2 + b^2 \geq 2ab\\\frac{a^2 + b^2}{ab} \geq \frac{2ab}{ab}\\\frac{a^2 + b^2}{ab} \geq 2\\\frac{a^2 + b^2}{ab}+2 \geq 4\\log_2(\frac{a^2 + b^2}{ab}+2) \geq log_2(4)\\log_2(\frac{a^2 + b^2+2ab}{ab}) \geq 2\\log_2[\frac{(a+b)*(a+b)}{ab}] \geq 2\\log_2[(a+b)*\frac{a+b}{ab}] \geq 2\\log_2(a+b)+log_2(\frac{a+b}{ab}) \geq 2\\log_2(a+b)+log_2(\frac{a}{ab}+\frac{b}{ab}) \geq 2\\log_2(a+b)+log_2(\frac{1}{b}+\frac{1}{a}) \geq 2$