Упр.468 ГДЗ Колягин Ткачёва 7 класс (Алгебра)
Выполнить умножение многочлена на одночлен:
1) $$\left(\frac{1}{2}a^3b^2-\frac{3}{4}ab^4\right)\frac{4}{3}a^3b$$;
2) $$\left(\frac{2}{3}a^2b^4+\frac{1}{2}a^3b\right)\frac{3}{2}ab^3$$;
3) $$\left(1\frac{4}{7}a^3x^3-2\frac{3}{4}a^2x^3-11ax^4\right)\left(-2\frac{6}{11}ax^6\right)$$;
4) $$\left(-2\frac{4}{9}b^6y+2\frac{1}{5}b^3y^2-11by^5\right)\left(-2\frac{1}{22}b^4y^5\right)$$.
Выполнить действия:
1) $$\frac{2}{x^2-9}+\frac{1}{x+3}$$;
2) $$\frac{5+p^2}{p^2-36}-\frac{p}{6+p}$$;
3) $$\frac{2x}{x-4}-\frac{5x-2}{x^2-16}$$.
$$\left(\frac12 a^3b^2-\frac34 ab^4\right)\frac43 a^3b$$
$$\frac12\cdot\frac43\,a^6b^3-\frac34\cdot\frac43\,a^4b^5=\frac23a^6b^3-ab^5.$$
$$\left(\frac23 a^2b^4+\frac12 a^3b\right)\frac32 ab^3$$
$$\frac23\cdot\frac32\,a^3b^7+\frac12\cdot\frac32\,a^4b^4=a^3b^7+\frac34a^4b^4.$$
$$\left(1\frac47 a^3x^3-2\frac34 a^2x^3-11ax^4\right)\left(-2\frac6{11}ax^6\right)$$
$$\left(\frac{11}{7}a^3x^3-\frac{11}{4}a^2x^3-11ax^4\right)\left(-\frac{28}{11}ax^6\right)$$
$$-\frac{28}{7}a^4x^9+\frac{28}{4}a^3x^9+28a^2x^{10}=-4a^4x^9+7a^3x^9+28a^2x^{10}.$$
$$\left(-2\frac49 b^6y+2\frac15 b^3y^2-11by^5\right)\left(-2\frac1{22}b^4y^5\right)$$
$$\left(-\frac{22}{9}b^6y+\frac{11}{5}b^3y^2-11by^5\right)\left(-\frac{45}{22}b^4y^5\right)$$
$$5b^{10}y^6-\frac{9}{2}b^7y^7+\frac{45}{2}b^5y^{10}.$$
$$\frac{2}{x^2-9}+\frac{1}{x+3}$$
$$x^2-9=(x-3)(x+3)$$
$$\frac{2}{(x-3)(x+3)}+\frac{1}{x+3}=\frac{2+x-3}{x^2-9}=\frac{x-1}{x^2-9}.$$
$$\frac{5+p^2}{p^2-36}-\frac{p}{6+p}$$
$$p^2-36=(p-6)(p+6)$$
$$\frac{5+p^2}{(p-6)(p+6)}-\frac{p}{p+6}=\frac{5+p^2-p(p-6)}{p^2-36}$$
$$\frac{5+p^2-p^2+6p}{p^2-36}=\frac{5+6p}{p^2-36}.$$
$$\frac{2x}{x-4}-\frac{5x-2}{x^2-16}$$
$$x^2-16=(x-4)(x+4)$$
$$\frac{2x(x+4)}{x^2-16}-\frac{5x-2}{x^2-16}=\frac{2x(x+4)-5x+2}{x^2-16}$$
$$\frac{2x^2+8x-5x+2}{x^2-16}=\frac{2x^2+3x+2}{x^2-16}.$$
Ответ
- $$\frac23a^6b^3-ab^5$$
- $$a^3b^7+\frac34a^4b^4$$
- $$-4a^4x^9+7a^3x^9+28a^2x^{10}$$
- $$5b^{10}y^6-\frac92b^7y^7+\frac{45}{2}b^5y^{10}$$
- $$\frac{x-1}{x^2-9}$$
- $$\frac{5+6p}{p^2-36}$$
- $$\frac{2x^2+3x+2}{x^2-16}$$
