Question

Решите аожалуйста номер 3.15 алгебра 7 класс абылкасымова. 25 баллов​

Answer 1

1)

$11^k:11^4*11^{k+1}=11^{k-4+k+1}=11^{2k-3\\}\\3)\\\frac{1}{4} ^{3k}:\frac{1}{4} ^k*\frac{1}{4} ^{2k+3}=\frac{1}{4} ^{3k-k+2k+3}=\frac{1}{4} ^{4k+3}\\5)\\(-\frac{1}{9}) ^{5t-2}:(-\frac{1}{9}) *(-\frac{1}{9}) ^{5t}=(-\frac{1}{9}) ^{5t-2-1+5t}=(-\frac{1}{9}) ^{10t-3}\\2) \\20^{10}:20^t*20^{3+t}=20^{10-t+3+t}=20^{13}\\4)\\(-9)^{20t}:(-9)^{t+5}:(-9)=(-9)^{20t-t-5-1}=(-9)^{19t-6}\\6)\\2.1^{k+3}*2.1^{6t}:2.1^{4t+3}=2.1^{k+3+6t-4t-3}=2.1^{k+2t}$

Answer 2

$\displaystyle\bf\\1)\\\\11^{k} :11^{4} \cdot 11^{k+1} =11^{k-4+k+1} =11^{2k-3} \\\\\\2)\\\\20^{10} :20^{t} \cdot 20^{3+t} =20^{10-t+3+t} =20^{13} \\\\\\3)\\\\\Bigg(\frac{1}{4} \Bigg)^{3k}: \Bigg(\frac{1}{4} \Bigg)^{k}\cdot\Bigg(\frac{1}{4} \Bigg)^{2k+3} =\Bigg(\frac{1}{4} \Bigg)^{3k-k+2k+3}=\Bigg(\frac{1}{4} \Bigg)^{4k+3}\\\\\\4)\\\\\Big(-9\Big)^{20t} :\Big(-9\Big)^{t+5} :\Big(-9\Big)=\Big(-9\Big)^{20t-t-5-1} =\Big(-9\Big)^{19t-6}$

$\displaystyle\bf\\5)\\\\\Bigg(-\frac{1}{9} \Bigg)^{5t-2}:\Bigg(-\frac{1}{9} \Bigg)\cdot \Bigg(-\frac{1}{9} \Bigg)^{5t}=\Bigg(-\frac{1}{9} \Bigg)^{5t-2-1+5t}=\Bigg(-\frac{1}{9} \Bigg)^{10t-3}\\\\\\6)\\\\2,1^{k+3} \cdot 2,1^{6t} :2,1^{4t+3} =2,1^{k+3+6t-4t-3} =2,1^{k+2t}$