Упр.23.12 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) (5a^0,4+b^0,2)(3a^0,4-4b^0,2); 5) (b^(4/3)-b^(-2/3))^2;
2) (m^0,5+n^0,5)(m^0,5-n^0,5); 6) (x^(1/6)+2)(x^(1/3)-2x^(1/6)+4);
3) (a^(1/3)-5b^(-1/4))(a^(1/3)+5b^(-1/4)); 7) (y^1,5-4y^0,5)+8y^2;
4) (m^(1/2)-n^(1/2))^2; 8) (a^(1/8)-1)(a^(1/4)+1)(a^(1/8)+1).

1. $$\begin{aligned} (5a^{0,4}+b^{0,2})(3a^{0,4}-4b^{0,2})&=15a^{0,8}-20a^{0,4}b^{0,2}+3a^{0,4}b^{0,2}-4b^{0,4}\\ &=15a^{0,8}-17a^{0,4}b^{0,2}-4b^{0,4}\\ &=15\sqrt[5]{a^4}-17\sqrt[5]{a^2b}-4\sqrt[5]{b^2}. \end{aligned}$$
2. $$\begin{aligned} (m^{0,5}+n^{0,5})(m^{0,5}-n^{0,5})&=m^{1}-n^{1}\\ &=m-n. \end{aligned}$$
3. $$\begin{aligned} \left(a^{\frac13}-5b^{-\frac14}\right)\left(a^{\frac13}+5b^{-\frac14}\right) &=a^{\frac23}-25b^{-\frac12}\\ &=\sqrt[3]{a^2}-\frac{25}{\sqrt{b}}. \end{aligned}$$
4. $$\begin{aligned} \left(m^{\frac12}-n^{\frac12}\right)^2&=m-2\sqrt{mn}+n. \end{aligned}$$
5. $$\begin{aligned} \left(b^{\frac43}-b^{-\frac23}\right)^2 &=b^{\frac83}-2b^{\frac23}+b^{-\frac43}\\ &=\sqrt[3]{b^8}-2\sqrt[3]{b^2}+\frac{1}{\sqrt[3]{b^4}}. \end{aligned}$$
6. $$\begin{aligned} \left(x^{\frac16}+2\right)\left(x^{\frac13}-2x^{\frac16}+4\right) &=x^{\frac12}+8\\ &=\sqrt{x}+8. \end{aligned}$$
7. $$\begin{aligned} \left(y^{1,5}-4y^{0,5}\right)^2+8y^2 &=y^3-8y^{1,5}y^{0,5}+16y+8y^2\\ &=y^3-8y^2+16y+8y^2\\ &=y^3+16y. \end{aligned}$$
8. $$\begin{aligned} \left(a^{\frac18}-1\right)\left(a^{\frac14}+1\right)\left(a^{\frac18}+1\right) &=\left(a^{\frac14}-1\right)\left(a^{\frac14}+1\right)\\ &=a^{\frac12}-1\\ &=\sqrt{a}-1. \end{aligned}$$

Ответ

1) $$15\sqrt[5]{a^4}-17\sqrt[5]{a^2b}-4\sqrt[5]{b^2}$$
2) $$m-n$$
3) $$\sqrt[3]{a^2}-\frac{25}{\sqrt{b}}$$
4) $$m-2\sqrt{mn}+n$$
5) $$\sqrt[3]{b^8}-2\sqrt[3]{b^2}+\frac{1}{\sqrt[3]{b^4}}$$
6) $$\sqrt{x}+8$$
7) $$y^3+16y$$
8) $$\sqrt{a}-1$$