Упр.42.30 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
- Докажите тождество:
- $$\left(\frac{m-n}{m^{\frac{3}{4}}+m^{\frac{1}{2}}n^{\frac{1}{4}}}-\frac{m^{\frac{1}{2}}-n^{\frac{1}{2}}}{m^{\frac{1}{4}}+n^{\frac{1}{4}}}\right)\left(\frac{n}{m}\right)^{-\frac{1}{2}}=m^{\frac{1}{4}}-n^{\frac{1}{4}};$$
- $$\frac{a+b}{a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}-\frac{a-b}{a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}-\frac{a^{\frac{2}{3}}-b^{\frac{2}{3}}}{a^{\frac{1}{3}}-b^{\frac{1}{3}}}=b^{\frac{1}{3}}-a^{\frac{1}{3}};$$
- $$\left(\frac{9}{a+8}-\frac{a^{\frac{1}{3}}+2}{a^{\frac{2}{3}}-2a^{\frac{1}{3}}+4}\right)\cdot\frac{a^{\frac{4}{3}}+8a^{\frac{1}{3}}}{1-a^{\frac{2}{3}}}+\frac{5-a^{\frac{2}{3}}}{1+a^{\frac{1}{3}}}=5;$$
- $$\left(\frac{m^{\frac{3}{4}}-n}{m^{\frac{1}{4}}-n^{\frac{1}{3}}}-3\left(m^3n^4\right)^{\frac{1}{12}}\right):\left(\frac{m^{\frac{3}{4}}+n}{m^{\frac{1}{4}}+n^{\frac{1}{3}}}-n^{\frac{2}{3}}\right)^2=\frac{1}{m^{\frac{1}{2}}}.$$
1)
$$\left(\frac{m-n}{m^{3/4}+m^{1/2}n^{1/4}}-\frac{m^{1/2}-n^{1/2}}{m^{1/4}+n^{1/4}}\right)\left(\frac{n}{m}\right)^{-1/2}$$
$$=\left(\frac{m-n}{m^{1/2}\left(m^{1/4}+n^{1/4}\right)}-\frac{m^{1/2}\left(m^{1/2}-n^{1/2}\right)}{m^{1/2}\left(m^{1/4}+n^{1/4}\right)}\right)\left(\frac{m}{n}\right)^{1/2}$$
$$=\frac{m-n-m+m^{1/2}n^{1/2}}{m^{1/2}\left(m^{1/4}+n^{1/4}\right)}\cdot \frac{m^{1/2}}{n^{1/2}} =\frac{m^{1/2}n^{1/2}-n}{n^{1/2}\left(m^{1/4}+n^{1/4}\right)}$$
$$=\frac{n^{1/2}\left(m^{1/2}-n^{1/2}\right)}{n^{1/2}\left(m^{1/4}+n^{1/4}\right)} =\frac{\left(m^{1/4}-n^{1/4}\right)\left(m^{1/4}+n^{1/4}\right)}{m^{1/4}+n^{1/4}} =m^{1/4}-n^{1/4}.$$
Тождество доказано.
2)
$$\frac{a+b}{a^{2/3}-a^{1/3}b^{1/3}+b^{2/3}} -\frac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}} -\frac{a^{2/3}-b^{2/3}}{a^{1/3}-b^{1/3}}$$
$$=\frac{(a+b)(a^{1/3}+b^{1/3})}{(a^{1/3}+b^{1/3})(a^{2/3}-a^{1/3}b^{1/3}+b^{2/3})} -\frac{(a-b)(a^{1/3}-b^{1/3})}{(a^{1/3}-b^{1/3})(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3})} -\frac{(a^{1/3}-b^{1/3})(a^{1/3}+b^{1/3})}{a^{1/3}-b^{1/3}}$$
$$=(a^{1/3}+b^{1/3})-(a^{1/3}-b^{1/3})-(a^{1/3}+b^{1/3}) =b^{1/3}-a^{1/3}.$$
Тождество доказано.
3)
$$\left(\frac{9}{a+8}-\frac{a^{1/3}+2}{a^{2/3}-2a^{1/3}+4}\right)\cdot \frac{a^{4/3}+8a^{1/3}}{1-a^{2/3}}+\frac{5-a^{2/3}}{1+a^{1/3}}$$
$$=\left(\frac{9}{(a^{1/3}+2)(a^{2/3}-2a^{1/3}+4)}-\frac{a^{1/3}+2}{a^{2/3}-2a^{1/3}+4}\right)\cdot \frac{a^{1/3}(a+8)}{1-a^{2/3}}+\frac{5-a^{2/3}}{1+a^{1/3}}$$
$$=\frac{9-(a^{1/3}+2)^2}{a+8}\cdot \frac{a^{1/3}(a+8)}{1-a^{2/3}}+\frac{5-a^{2/3}}{1+a^{1/3}}$$
$$=\frac{5-a^{2/3}-4a^{1/3}}{1-a^{2/3}}\cdot a^{1/3}+\frac{5-a^{2/3}}{1+a^{1/3}}$$
$$=\frac{a^{1/3}(5-a^{2/3}-4a^{1/3})(1+a^{1/3})+(5-a^{2/3})(1-a^{2/3})}{(1-a^{2/3})(1+a^{1/3})} =5.$$
Тождество доказано.
4)
$$\left(\frac{m^{3/4}-n}{m^{1/4}-n^{1/3}}-3\sqrt[12]{m^3n^4}\right):\left(\frac{m^{3/4}+n}{m^{1/4}+n^{1/3}}-n^{2/3}\right)^2$$
$$=\left(\frac{m^{3/4}-n}{m^{1/4}-n^{1/3}}-3m^{1/4}n^{1/3}\right):\left(\frac{m^{3/4}+n-n^{2/3}(m^{1/4}+n^{1/3})}{m^{1/4}+n^{1/3}}\right)^2$$
$$=\frac{m^{3/4}-n-3m^{1/4}n^{1/3}(m^{1/4}-n^{1/3})}{m^{1/4}-n^{1/3}} :\left(\frac{m^{3/4}-n^{2/3}m^{1/4}}{m^{1/4}+n^{1/3}}\right)^2$$
$$=\frac{(m^{1/4}-n^{1/3})^3}{m^{1/4}-n^{1/3}} :\left(\frac{m^{1/4}(m^{1/2}-n^{2/3})}{m^{1/4}+n^{1/3}}\right)^2$$
$$=(m^{1/4}-n^{1/3})^2\cdot \frac{(m^{1/4}+n^{1/3})^2}{m^{1/2}(m^{1/2}-n^{2/3})^2} =\frac{1}{m^{1/2}}.$$
Тождество доказано.
