Упр.13.44 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1. 13.44. Упростите выражение:
   1) $$0{,}42ac^3\cdot1\frac{3}{7}a^4c^2$$;
   2) $$1{,}2xyz\cdot2\frac{1}{6}x^5y^6$$;
   3) $$-2\frac{1}{3}m^2np^3\cdot\left(\frac{3}{7}np^4\right)^2$$;
   4) $$\left(1\frac{1}{2}x^2y^3\right)^5\cdot\frac{16}{27}x^8y^2$$.

1. $$0{,}42ac^3 \cdot 1\frac{3}{7}a^4c^2=0{,}42\cdot \frac{10}{7}\cdot a^{1+4}c^{3+2}$$
   $$0{,}42\cdot \frac{10}{7}=0{,}6$$
   $$0{,}42ac^3 \cdot 1\frac{3}{7}a^4c^2=0{,}6a^5c^5$$
2. $$1{,}2xyz \cdot 2\frac{1}{6}x^5y^6=1{,}2\cdot \frac{13}{6}\cdot x^{1+5}y^{1+6}z$$
   $$1{,}2\cdot \frac{13}{6}=2{,}6$$
   $$1{,}2xyz \cdot 2\frac{1}{6}x^5y^6=2{,}6x^6y^7z$$
3. $$-2\frac{1}{3}m^2np^3\cdot \left(\frac{3}{7}np^4\right)^2=-\frac{7}{3}m^2np^3\cdot \frac{9}{49}n^2p^8$$
   $$-\frac{7}{3}\cdot \frac{9}{49}=-\frac{3}{7}$$
   $$-2\frac{1}{3}m^2np^3\cdot \left(\frac{3}{7}np^4\right)^2=-\frac{3}{7}m^2n^3p^{11}$$
4. $$\left(1\frac{1}{2}x^2y^3\right)^5\cdot \frac{16}{27}x^8y^2=\left(\frac{3}{2}\right)^5x^{10}y^{15}\cdot \frac{16}{27}x^8y^2$$
   $$\left(\frac{3}{2}\right)^5=\frac{243}{32}$$
   $$\frac{243}{32}\cdot \frac{16}{27}=\frac{9}{2}$$
   $$\left(1\frac{1}{2}x^2y^3\right)^5\cdot \frac{16}{27}x^8y^2=\frac{9}{2}x^{18}y^{17}=4{,}5x^{18}y^{17}$$

Ответ

1) $$0{,}6a^5c^5$$; 2) $$2{,}6x^6y^7z$$; 3) $$-\frac{3}{7}m^2n^3p^{11}$$; 4) $$4{,}5x^{18}y^{17}$$.