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bạn hãy nhân ở mẫu với biểu thức tương ướng để tạo ra biểu thức liên hợp , là HĐT số 3 ạ
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..
\(\hept{\begin{cases}\sqrt[3]{3}=a\\\sqrt[3]{4}=b\end{cases}}\)
\(\Rightarrow b^3-a^3=1\)
\(\Leftrightarrow-b^2-ab=a^2+\frac{1}{a-b}\)
Ta cần trục cái:
\(\frac{1}{a^2-ab-b^2}=\frac{1}{a^2+a^2+\frac{1}{a-b}}=\frac{a-b}{2a^3-2a^2b+1}\)
\(=\frac{\sqrt[3]{3}-\sqrt[3]{4}}{7-2\sqrt[3]{36}}=\frac{\left(\sqrt[3]{3}-\sqrt[3]{4}\right)\left(49+14\sqrt[3]{36}+24\sqrt[3]{6}\right)}{55}=\frac{\sqrt[3]{3}-7\sqrt[3]{4}-4\sqrt[3]{18}}{55}\)
Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)
a; \(=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{3+2}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{5}\)
b; tương tự
\(A=\frac{1}{\sqrt[3]{9}-\sqrt[3]{3}+\sqrt[3]{24}-\sqrt[3]{243}+\sqrt[3]{375}}\)
\(=\frac{1}{\sqrt[3]{9}-\sqrt[3]{3}+\sqrt[3]{8.3}-\sqrt[3]{27.9}+\sqrt[3]{125.3}}\)
\(=\frac{1}{\sqrt[3]{9}-\sqrt[3]{3}+2\sqrt[3]{3}-3\sqrt[3]{9}+5\sqrt[3]{3}}\)
\(=\frac{1}{6\sqrt[3]{3}-2\sqrt[3]{9}}=\frac{1}{2\sqrt[3]{9}.\left(\sqrt[3]{9}-1\right)}\)
\(=\frac{\sqrt[3]{81}.\left(\sqrt[3]{81}+\sqrt[3]{9}+1\right)}{2\sqrt[3]{9}.\left(\sqrt[3]{9}-1\right)\left(\sqrt[3]{81}+\sqrt[3]{9}+1\right).\sqrt[3]{81}}\)
\(=\frac{9\sqrt[3]{9}+9+3\sqrt[3]{3}}{144}\)
p/s: mk k chắc, sai đâu mn ib cho mk nhé
thank bạn nhiều nha