Упр.938 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б (77^3-69^3)/(70^2-62^2) — (77^3+41^3)/(125^2-49)-1/2;
в) (65^2-32^2-97*11)/(61^2-36^2) + (56^2-26^2)/(66^2-16^2);
г) (109^2+160*32 — 51^2)/(139^2-11^2) + (42^2-36)/(84^2-12^2).

Подробный ответ

а)

$$ \frac12+\frac{53^2-27^2}{31^2-25}-\frac{53^2-27^2}{58^2-22^2} = \frac12+\frac{(53-27)(53+27)}{(31-5)(31+5)}-\frac{(53-27)(53+27)}{(58-22)(58+22)} $$
$$ = \frac12+\frac{26\cdot 80}{26\cdot 36}-\frac{26\cdot 80}{36\cdot 80} = \frac12+\frac{20}{9}-\frac{13}{18} = \frac{9+40-13}{18} = \frac{36}{18} =2. $$

б)

$$ \frac{77^3-69^3}{70^2-62^2}-\frac{77^3+41^3}{125^2-49}-\frac12 = \frac{(77-69)(77^2+77\cdot 69+69^2)}{(70-62)(70+62)} -\frac{(77+41)(77^2-77\cdot 41+41^2)}{(125-7)(125+7)}-\frac12 $$
$$ = \frac{8(77^2+77\cdot 69+69^2)}{8\cdot 132} -\frac{118(77^2-77\cdot 41+41^2)}{118\cdot 132} -\frac12 $$
$$ = \frac{77^2+77\cdot 69+69^2}{132} -\frac{77^2-77\cdot 41+41^2}{132} -\frac12 $$
$$ = \frac{77\cdot 69+77\cdot 41+69^2-41^2}{132}-\frac12 = \frac{77(69+41)+(69-41)(69+41)}{132}-\frac12 $$
$$ = \frac{110\cdot 105+28\cdot 110}{132}-\frac12 = \frac{11550+3080}{132}-\frac12 = \frac{14630}{132}-\frac12 = 87. $$

в)

$$ \frac{65^2-32^2-97\cdot 11}{61^2-36^2}+\frac{56^2-26^2}{66^2-16^2} = \frac{(65-32)(65+32)-97\cdot 11}{(61-36)(61+36)} +\frac{(56-26)(56+26)}{(66-16)(66+16)} $$
$$ = \frac{33\cdot 97-97\cdot 11}{25\cdot 97}+\frac{30\cdot 82}{50\cdot 82} = \frac{97(33-11)}{25\cdot 97}+\frac35 = \frac{22}{25}+\frac35 = \frac{22}{25}+\frac{15}{25} = \frac{37}{25} =1{,}48. $$

г)

$$ \frac{109^2+160\cdot 32-51^2}{139^2-11^2}+\frac{42^2-36}{84^2-12^2} = \frac{(109-51)(109+51)+160\cdot 32}{(139-11)(139+11)} +\frac{42^2-6^2}{(84-12)(84+12)} $$
$$ = \frac{58\cdot 160+160\cdot 32}{128\cdot 150} +\frac{(42-6)(42+6)}{72\cdot 96} = \frac{160(58+32)}{128\cdot 150} +\frac{36\cdot 48}{72\cdot 96} $$
$$ = \frac{160\cdot 90}{128\cdot 150}+\frac{1}{4} = \frac{3}{4}+\frac{1}{4} =1. $$