Упр.185 ГДЗ Никольский Потапов 8 класс (Алгебра)

Задача

а) 3; б) 5; в) 6; г) 7; д) 8; е) 10; ж) 11; з) 12.

Подробный ответ

Найдём приближённые значения корней, последовательно уточняя их по квадратам соседних десятичных чисел.

а) $$\sqrt{3}\approx 1{,}73$$

$$1{,}7^2=2{,}89,\quad 1{,}8^2=3{,}24$$
$$1{,}7<\sqrt{3}<1{,}8$$
$$1{,}72^2=2{,}96,\quad 1{,}73^2=2{,}99,\quad 1{,}74^2=3{,}03$$
$$1{,}73<\sqrt{3}<1{,}74$$

б) $$\sqrt{5}\approx 2{,}23$$

$$2{,}2^2=4{,}84,\quad 2{,}3^2=5{,}29$$
$$2{,}2<\sqrt{5}<2{,}3$$
$$2{,}22^2=4{,}93,\quad 2{,}23^2=4{,}97,\quad 2{,}24^2=5{,}02$$
$$2{,}23<\sqrt{5}<2{,}24$$

в) $$\sqrt{6}\approx 2{,}45$$

$$2{,}4^2=5{,}76,\quad 2{,}5^2=6{,}25$$
$$2{,}4<\sqrt{6}<2{,}5$$
$$2{,}43^2=5{,}90,\quad 2{,}44^2=5{,}95,\quad 2{,}45^2=6{,}0025$$
$$2{,}44<\sqrt{6}<2{,}45$$

г) $$\sqrt{7}\approx 2{,}65$$

$$2{,}6^2=6{,}76,\quad 2{,}7^2=7{,}29$$
$$2{,}6<\sqrt{7}<2{,}7$$
$$2{,}63^2=6{,}9169,\quad 2{,}64^2=6{,}9696,\quad 2{,}65^2=7{,}0225$$
$$2{,}64<\sqrt{7}<2{,}65$$

д) $$\sqrt{8}\approx 2{,}83$$

$$2{,}8^2=7{,}84,\quad 2{,}9^2=8{,}41$$
$$2{,}8<\sqrt{8}<2{,}9$$
$$2{,}81^2=7{,}8961,\quad 2{,}82^2=7{,}9524,\quad 2{,}83^2=8{,}0089$$
$$2{,}82<\sqrt{8}<2{,}83$$

е) $$\sqrt{10}\approx 3{,}16$$

$$3^2=9,\quad 4^2=16$$
$$3<\sqrt{10}<4$$
$$3{,}1^2=9{,}61,\quad 3{,}2^2=10{,}24$$
$$3{,}1<\sqrt{10}<3{,}2$$
$$3{,}15^2=9{,}9225,\quad 3{,}16^2=9{,}9856,\quad 3{,}17^2=10{,}0489$$
$$3{,}16<\sqrt{10}<3{,}17$$

ж) $$\sqrt{11}\approx 3{,}32$$

$$3^2=9,\quad 4^2=16$$
$$3<\sqrt{11}<4$$
$$3{,}3^2=10{,}89,\quad 3{,}4^2=11{,}56$$
$$3{,}3<\sqrt{11}<3{,}4$$
$$3{,}31^2=10{,}9561,\quad 3{,}32^2=11{,}0224$$
$$3{,}31<\sqrt{11}<3{,}32$$

з) $$\sqrt{12}\approx 3{,}46$$

$$3^2=9,\quad 4^2=16$$
$$3<\sqrt{12}<4$$
$$3{,}4^2=11{,}56,\quad 3{,}5^2=12{,}25$$
$$3{,}4<\sqrt{12}<3{,}5$$
$$3{,}45^2=11{,}9025,\quad 3{,}46^2=11{,}9716,\quad 3{,}47^2=12{,}0409$$
$$3{,}46<\sqrt{12}<3{,}47$$