Question

Помогите умоляюю ((хоть 4 задания плз

Answer 1

(1)
a)
<u />$0.4\sqrt{10}*\sqrt{250}+\sqrt{169}=0.4*\sqrt{2500}+13}=0.4*50+13=\boxed{33}$

б)
$\sqrt{24}-4\sqrt{6}+\sqrt{54}=\sqrt{6*4}-4\sqrt{6}+\sqrt{6*9}=\sqrt{6}(\sqrt{4}-4+\sqrt{9})=\\=\sqrt{6}(2-4+3)=\boxed{\sqrt{6}}$





(3)
$\frac{4-c}{c+2\sqrt{c}}=\frac{2^2-(\sqrt{c})^2}{(\sqrt{c})^2+2\sqrt{c}}=\frac{(2-\sqrt{c})(2+\sqrt{c})}{\sqrt{c}(\sqrt{c}+2)}=\frac{2-\sqrt{c}}{\sqrt{c}}=\boxed{\frac{2}{\sqrt{c}}-1}=\frac{2\sqrt{c}}{c}-1$





(4)
$B=\sqrt{0.17^2-0.08^2}=\sqrt{(0.17-0.08)(0.17+0.08)}=\sqrt{0.9*0.25}=\\=0.3*0.5=0.15\\A=0.(15)=0.15151515...\\A>B$




(5)
$\frac{\sqrt{15}+4}{4-\sqrt{15}}=31+8\sqrt{15}\\\sqrt{15}+4=(31+8\sqrt{15})(4-\sqrt{15})\\\sqrt{15}+4=124-31\sqrt{15}+32\sqrt{15}-8*15\\\sqrt{15}+4=\sqrt{15}+4$
Доказано.