Question

se considera a=1/radical2+2/radical 8+ 3radical 18+4radical 32 si b = radical 13^2-5^2 totul supra radical 10^2-8^2.Aratati va a^2-b^2

Answer 1

$a=\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{8}}+\frac{3}{\sqrt{18}}+\frac{4}{\sqrt{32}}$
$a=\frac{1}{\sqrt{2}}+\frac{2}{2\sqrt{2}}+\frac{3}{3\sqrt{2}}+\frac{4}{4\sqrt{2}}$
$a=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$
$a=\frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}$
$a^2=(2\sqrt{2})^2=4\cdot2=8$
$b=\frac{\sqrt{13^2-5^2}}{\sqrt{10^2-8^2}}$
$b=\frac{\sqrt{169-25}}{\sqrt{100-64}}$
$b=\frac{\sqrt{144}}{\sqrt{36}}$
$b=\frac{12}{6}=2$
$b^2=2^2=4$
$a^2-b^2=8-4=4$