Question

Вычислите помогите жппжпжпжпжппж​

Answer 1

Ответ:

$1)5\sqrt{2}$
$2)\frac{2\sqrt{55} }{{11}}$
$3)10$
$4)20$

Объяснение:

$1)\sqrt{\frac{(63^{4}-62^{4})*2}{(63^{2}+62^{2})*5} }=\sqrt{\frac{(63^{2}-62^{2})(63^{2}+62^{2})*2}{(63^{2}+62^{2})*5} }=\sqrt{\frac{(63^{2}-62^{2})*2}{5} }=\sqrt{\frac{(63-62)(63+62)*2}{5} }=\sqrt{\frac{125*2}{5} } =\sqrt{25*2}=5\sqrt{2}$
$2)\sqrt{\frac{86^{4}-74^{4}}{1,1*(32^{2}-8^{2})(86^{2}+74^{2})} }=\sqrt{\frac{(86^{2}-74^{2})(86^{2}+74^{2})}{1,1*(32^{2}-8^{2})(86^{2}+74^{2})} }=\sqrt{\frac{86^{2}-74^{2}}{1,1*(32^{2}-8^{2})} }=\sqrt{\frac{(86-74)(86+74)}{1,1*(32-8)(32+8)} }=\sqrt{\frac{12*160}{1,1*24*40} }=\sqrt{\frac{4}{1,1*2} }=\sqrt{\frac{2}{1,1} }=\frac{\sqrt{2} }{\sqrt{1,1}}=\frac{\sqrt{2,2} }{{1,1}}=\frac{\sqrt{220} }{{11}}=\frac{2\sqrt{55} }{{11}}$
$3)\sqrt{\frac{76^{4}-74^{4}}{3*(76^{2}+74^{2})} } =\sqrt{\frac{(76^{2}-74^{2})(76^{2}+74^{2})}{3*(76^{2}+74^{2})} }=\sqrt{\frac{76^{2}-74^{2}}{3}}=\sqrt{\frac{(76-74)(76+74)}{3}}=\sqrt{\frac{2*150}{3}}=\sqrt{100}=10$
$4)\sqrt{\frac{(97^{4}-53^{4})*10}{(97^{2}+53^{2}(83^{2}-82^{2})} } =\sqrt{\frac{(97^{2}-53^{2})(97^{2}+53^{2})*10}{(97^{2}+53^{2})(83^{2}-82^{2})} }=\sqrt{\frac{(97^{2}-53^{2})*10}{83^{2}-82^{2}}}=\sqrt{\frac{(97-53)(97+53)*10}{(83-82)(83+82)}}=\sqrt{\frac{44*150*10}{165} } =\sqrt{\frac{30*44*10}{33} } =\sqrt{\frac{30*4*10}{3} } =\sqrt{10*4*10}=10*2=20$