Упр.947 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

947 а) ((a+b)/(a-) + (a-b)/(a+b)):(a2/(a2-b2) + 1/(a2/b2 — 1));
б) ((x2y-xy2)/(x-y) + xy)*(y/x+x/y);
в) (n/(m-n) +m/(m+n))* (m2/n2+n2/m2 — 2);
г) ((x+1)/(x-1) + (x-1)/(x+1) + 4x)*(x-1/x);
д) (1+a/x + a2/x2)(1-a/x) * x3/(a3-x3);
е) (4x-3)/(3-2x) — (4+5x)/(3+2x) — (3+x-10×2)/(4×2-9).

Подробный ответ

а)
$$\left(\frac{a+b}{a-b}+\frac{a-b}{a+b}\right):\left(\frac{a^2}{a^2-b^2}+\frac{1}{\frac{a^2}{b^2}-1}\right)=$$
$$=\frac{(a+b)^2+(a-b)^2}{(a-b)(a+b)}:\left(\frac{a^2}{a^2-b^2}+\frac{b^2}{a^2-b^2}\right)=$$
$$=\frac{2a^2+2b^2}{a^2-b^2}:\frac{a^2+b^2}{a^2-b^2}=$$
$$=\frac{2(a^2+b^2)}{a^2-b^2}\cdot\frac{a^2-b^2}{a^2+b^2}=2.$$
б)
$$\left(\frac{x^2y-xy^2}{x-y}+xy\right)\left(\frac{y}{x}+\frac{x}{y}\right)=$$
$$=\left(\frac{xy(x-y)}{x-y}+xy\right)\cdot\frac{x^2+y^2}{xy}=$$
$$=(xy+xy)\cdot\frac{x^2+y^2}{xy}=$$
$$=2xy\cdot\frac{x^2+y^2}{xy}=2(x^2+y^2).$$
в)
$$\left(\frac{n}{m-n}+\frac{m}{m+n}\right)\left(\frac{m^2}{n^2}+\frac{n^2}{m^2}-2\right)=$$
$$=\frac{n(m+n)+m(m-n)}{(m-n)(m+n)}\cdot\frac{m^4+n^4-2m^2n^2}{m^2n^2}=$$
$$=\frac{m^2+n^2}{m^2-n^2}\cdot\frac{(m^2-n^2)^2}{m^2n^2}=$$
$$=\frac{(m^2+n^2)(m^2-n^2)}{m^2n^2}=\frac{m^4-n^4}{m^2n^2}.$$
г)
$$\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+4x\right)\left(x-\frac{1}{x}\right)=$$
$$=\left(\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}+4x\right)\cdot\frac{x^2-1}{x}=$$
$$=\left(\frac{4x}{x^2-1}+4x\right)\cdot\frac{x^2-1}{x}=$$
$$=\frac{4x+4x(x^2-1)}{x^2-1}\cdot\frac{x^2-1}{x}=\frac{4x^3}{x}=4x^2.$$
д)
$$\left(1+\frac{a}{x}+\frac{a^2}{x^2}\right)\left(1-\frac{a}{x}\right)\cdot\frac{x^3}{a^3-x^3}=$$
$$=\frac{x^2+ax+a^2}{x^2}\cdot\frac{x-a}{x}\cdot\frac{x^3}{a^3-x^3}.$$
Так как $$a^3-x^3=(a-x)(a^2+ax+x^2),$$ то
$$=\frac{x^2+ax+a^2}{x^2}\cdot\frac{x-a}{x}\cdot\frac{x^3}{(a-x)(a^2+ax+x^2)}=-1.$$
е)
$$\frac{4x-3}{3-2x}-\frac{4+5x}{3+2x}-\frac{3+x-10x^2}{4x^2-9}=$$
$$=\frac{4x-3}{3-2x}-\frac{4+5x}{3+2x}+\frac{3+x-10x^2}{9-4x^2}.$$
Приведём к общему знаменателю:
$$=\frac{(4x-3)(3+2x)-(4+5x)(3-2x)+3+x-10x^2}{(3-2x)(3+2x)}.$$
Числитель:
$$12x+8x^2-9-6x-(12-8x+15x-10x^2)+3+x-10x^2=$$
$$=8x^2-18.$$
Тогда
$$\frac{8x^2-18}{(3-2x)(3+2x)}=\frac{2(4x^2-9)}{9-4x^2}=-2.$$