Упр.64 ГДЗ Алимов 10-11 класс (Алгебра)
Алимов, Колягин, Ткачёва
10 класс
Автор
Алимов, Колягин, Ткачёва
Упр.64 ГДЗ Алимов 10-11 класс (Алгебра)
Задача
1) a1/2 — b1/2;
2) y2/3 -1;
3) a1/3-b1/3;
4) x-y;
5) 4a1/2 — b1/2;
6) 0,01m1/6-n1/6.
Подробный ответ
- $$a^{\frac12}-b^{\frac12}= \left(a^{\frac14}\right)^2-\left(b^{\frac14}\right)^2$$
$$=\left(a^{\frac14}+b^{\frac14}\right)\left(a^{\frac14}-b^{\frac14}\right)$$ - $$y^{\frac23}-1=\left(y^{\frac13}\right)^2-1^2$$
$$=\left(y^{\frac13}+1\right)\left(y^{\frac13}-1\right)$$ - $$a^{\frac13}-b^{\frac13}=\left(a^{\frac16}\right)^2-\left(b^{\frac16}\right)^2$$
$$=\left(a^{\frac16}+b^{\frac16}\right)\left(a^{\frac16}-b^{\frac16}\right)$$ - $$x-y=x^1-y^1=\left(x^{\frac12}\right)^2-\left(y^{\frac12}\right)^2$$
$$=\left(x^{\frac12}+y^{\frac12}\right)\left(x^{\frac12}-y^{\frac12}\right)$$ - $$4a^{\frac12}-b^{\frac12}=\left(2a^{\frac14}\right)^2-\left(b^{\frac14}\right)^2$$
$$=\left(2a^{\frac14}+b^{\frac14}\right)\left(2a^{\frac14}-b^{\frac14}\right)$$ - $$0{,}01m^{\frac16}-n^{\frac16}=\left(0{,}1m^{\frac1{12}}\right)^2-\left(n^{\frac1{12}}\right)^2$$
$$=\left(0{,}1m^{\frac1{12}}+n^{\frac1{12}}\right)\left(0{,}1m^{\frac1{12}}-n^{\frac1{12}}\right)$$
Ответ
1) $$\left(a^{\frac14}+b^{\frac14}\right)\left(a^{\frac14}-b^{\frac14}\right)$$
2) $$\left(y^{\frac13}+1\right)\left(y^{\frac13}-1\right)$$
3) $$\left(a^{\frac16}+b^{\frac16}\right)\left(a^{\frac16}-b^{\frac16}\right)$$
4) $$\left(x^{\frac12}+y^{\frac12}\right)\left(x^{\frac12}-y^{\frac12}\right)$$
5) $$\left(2a^{\frac14}+b^{\frac14}\right)\left(2a^{\frac14}-b^{\frac14}\right)$$
6) $$\left(0{,}1m^{\frac1{12}}+n^{\frac1{12}}\right)\left(0{,}1m^{\frac1{12}}-n^{\frac1{12}}\right)$$
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