Упр.38.17 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (a^4-1)/(a^3-a)·a/(1+a^2 );
2) (a^2-8ab)/12b : (8b^2-ab)/24a;
3) (5m^2-5n^2)/(m^2+n^2 ) : (15n-15m)/(4m^2+4n^2 );
4) (mn^2-36m)/(m^3-8) : (2n+12)/(6m-12);
5) (a^4-1)/(a^2-a+1) : (a-1)/(a^3+1);
6) (4x^2-100)/6x : (2x^2-20x+50).

1. $$\frac{a^4-1}{a^3-a}\cdot \frac{a}{1+a^2}= \frac{(a^2-1)(a^2+1)}{a(a^2-1)}\cdot \frac{a}{1+a^2}=1.$$

2. $$\frac{a^2-8ab}{12b}:\frac{8b^2-ab}{24a}= \frac{a(a-8b)}{12b}:\frac{b(8b-a)}{24a}$$
   $$=\frac{a(a-8b)}{12b}\cdot \frac{24a}{b(8b-a)}= \frac{2a^2(a-8b)}{b^2(8b-a)}=-\frac{2a^2}{b^2}.$$

3. $$\frac{5m^2-5n^2}{m^2+n^2}:\frac{15n-15m}{4m^2+4n^2}= \frac{5(m^2-n^2)}{m^2+n^2}:\frac{15(n-m)}{4(m^2+n^2)}$$
   $$=\frac{5(m-n)(m+n)}{m^2+n^2}\cdot \frac{4(m^2+n^2)}{15(n-m)}= \frac{4(m-n)(m+n)}{3(n-m)}=-\frac{4(m+n)}{3}.$$

4. $$\frac{mn^2-36m}{m^3-8}:\frac{2n+12}{6m-12}= \frac{m(n^2-36)}{m^3-8}:\frac{2(n+6)}{6(m-2)}$$
   $$=\frac{m(n-6)(n+6)}{(m-2)(m^2+2m+4)}\cdot \frac{3(m-2)}{n+6} =\frac{3m(n-6)}{m^2+2m+4}.$$

5. $$\frac{a^4-1}{a^2-a+1}:\frac{a-1}{a^3+1}= \frac{(a^2-1)(a^2+1)}{a^2-a+1}:\frac{a-1}{(a+1)(a^2-a+1)}$$
   $$=\frac{(a-1)(a+1)(a^2+1)}{a^2-a+1}\cdot \frac{(a+1)(a^2-a+1)}{a-1}=(a+1)^2(a^2+1).$$

6. $$\frac{4x^2-100}{6x}:(2x^2-20x+50)= \frac{4(x^2-25)}{6x}:(2(x^2-10x+25))$$
   $$=\frac{4(x-5)(x+5)}{6x}:2(x-5)^2 =\frac{4(x-5)(x+5)}{6x}\cdot \frac{1}{2(x-5)^2} =\frac{x+5}{3x(x-5)}.$$

Ответ

1) $$1$$;
2) $$-\frac{2a^2}{b^2}$$;
3) $$-\frac{4(m+n)}{3}$$;
4) $$\frac{3m(n-6)}{m^2+2m+4}$$;
5) $$ (a+1)^2(a^2+1) $$;
6) $$\frac{x+5}{3x(x-5)}$$.