Т.к <span>x=ab+1,y=ab-1,то
ab+1-(ab-1)-1=ab+1-ab+1-1=1
</span>
<span>(1-х)(х+1)+(х-1)</span>
![\sqrt[7]{b^3}\sqrt[4]{b} =b^{\frac{3}{7}}b^{\frac{1}{4}}=b^{\frac{3}{7}+\frac{1}{4}}](https://tex.z-dn.net/?f=+%5Csqrt%5B7%5D%7Bb%5E3%7D%5Csqrt%5B4%5D%7Bb%7D+%3Db%5E%7B%5Cfrac%7B3%7D%7B7%7D%7Db%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3Db%5E%7B%5Cfrac%7B3%7D%7B7%7D%2B%5Cfrac%7B1%7D%7B4%7D%7D)
Подсчитаем отдельно степень. Приведем к общему знаменателю
![\frac{3}{7}+\frac{1}{4}}=\frac{3*4}{28}+\frac{1*7}{28}=\frac{12+7}{28}=\frac{19}{28}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B7%7D%2B%5Cfrac%7B1%7D%7B4%7D%7D%3D%5Cfrac%7B3%2A4%7D%7B28%7D%2B%5Cfrac%7B1%2A7%7D%7B28%7D%3D%5Cfrac%7B12%2B7%7D%7B28%7D%3D%5Cfrac%7B19%7D%7B28%7D)
Значит
![b^{\frac{3}{7}+\frac{1}{4}}=b^{\frac{19}{28}}](https://tex.z-dn.net/?f=b%5E%7B%5Cfrac%7B3%7D%7B7%7D%2B%5Cfrac%7B1%7D%7B4%7D%7D%3Db%5E%7B%5Cfrac%7B19%7D%7B28%7D%7D)
Ответ: с)
![b^{\frac{19}{28}}](https://tex.z-dn.net/?f=b%5E%7B%5Cfrac%7B19%7D%7B28%7D%7D)
Объяснение:
(a-3b)/b=4
a-3b=4b
a=4b+3b=7b
a/b=7b/b=7
(4a+5b)/a=(4•7b+5b)/(7b)=(28b+5b)/(7b)=33b/(7b)=33/7=4 5/7