Задание 623

Решение:

• \({\largeа)}\ \frac{a^3-b^3}{a^4-b^4}=\frac{ (a-b)(a^2+ab+b^2)}{ (a-b)(a^3+a^2b+ab^2+b^3)}=\frac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}\)

• \({\largeб)}\ \frac{a^3+b^3}{a^2-ab+b^2}=\frac{ (a+b)(a^2-ab+b^2)}{a^2-ab+b^2}=a+b\)

• \({\largeв)}\ \frac{a^5-b^5}{a^3-b^3}=\frac{ (a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)}{ (a-b)(a^2+ab+b^2)}=\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\)

• \({\largeг)}\ \frac{a^5+b^5}{a^7+b^7}=\frac{ (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)}{ (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)}=\frac{a^4-a^3b+a^2b^2-ab^3+b^4}{a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6}\)

• \({\largeд)}\ \frac{a^3+a^2b+ab^2+b^3}{a^4-b^4}=\frac{a^3+a^2b+ab^2+b^3}{ (a-b)(a^3+a^2b+ab^2+b^3)}=\frac{1}{a-b}\)

• \({\largeе)}\ \frac{a^5+b^5}{a^4-a^3b+a^2b^2-ab^3+b^4}=\frac{ (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)}{a^4-a^3b+a^2b^2-ab^3+b^4}=a+b\)

• \({\largeж)}\ \frac{a^3-8}{a^4-16}=\frac{a^3-2^3}{a^4-2^4}=\frac{ (a-2)(a^2+2a+2^2)}{ (a-2)(a^3+2a^2+2^2a+2^3)}=\frac{ (a-2)(a^2+2a+4)}{ (a-2)(a^3+2a^2+4a+8)}=\frac{a^2+2a+4}{a^3+2a^2+4a+8}\)

• \({\largeз)}\ \frac{a^3+27}{a^2-3a+9}=\frac{a^3+3^3}{a^2-3a+9}=\frac{ (a+3)(a^2-3a+3^2)}{a^2-3a+9}=\frac{ (a+3)(a^2-3a+9)}{a^2-3a+9}=a+3\)

• \({\largeи)}\ \frac{a^5-32}{a^3-8}=\frac{a^5-2^5}{a^3-2^3}=\frac{ (a-2)(a^4+2a^3+2^2a^2+2^3a+2^4)}{ (a-2)(a^2+2a+2^2)}=\frac{ (a-2)(a^4+2a^3+4a^2+8a+16)}{ (a-2)(a^2+2a+4)}=\frac{a^4+2a^3+4a^2+8a+16}{a^2+2a+4}\)

• \({\largeк)}\ \frac{a^5+32}{a^7+128}=\frac{a^5+2^5}{a^7+2^7}=\frac{ (a+2)(a^4-2a^3+2^2a^2-2^3a+2^4)}{ (a+2)(a^6-2a^5+2^2a^4-2^3a^3+2^4a^2-2^5a+2^6)}=\frac{ (a+2)(a^4-2a^3+4a^2-8a+16)}{ (a+2)(a^6-2a^5+4a^4-8a^3+16a^2-32a+64)}=\frac{a^4-2a^3+4a^2-8a+16}{a^6-2a^5+4a^4-8a^3+16a^2-32a+64}\)

• \({\largeл)}\ \frac{a^3+2a^2+4a+8}{a^4-16}=\frac{a^3+2a^2+4a+8}{a^4-2^4}=\frac{a^3+2a^2+4a+8}{ (a-2)(a^3+2a^2+2^2a+2^3)}=\frac{a^3+2a^2+4a+8}{ (a-2)(a^3+2a^2+4a+8)}=\frac{1}{a-2}\)

• \({\largeм)}\ \frac{a^5+1}{a^4-a^3+a^2-a+1}=\frac{ (a+1)(a^4-a^3+a^2-a+1)}{a^4-a^3+a^2-a+1}=a+1\)