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a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-5-2\sqrt{6}}\)
\(=\dfrac{-\sqrt{5}-\sqrt{3}-\sqrt{2}}{2\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)}{12}\)
b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)
\(=\dfrac{2\left(-1-\sqrt{2}-\sqrt{3}\right)}{\left(-1-\sqrt{2}\right)^2-3}\)
\(=\dfrac{\left(-1-\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{-\sqrt{2}-2-\sqrt{6}}{2}\)
Lời giải:
a.
\(\frac{1}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{\sqrt{5}+\sqrt{3}}{5-3}=\frac{\sqrt{5}+\sqrt{3}}{2}\)
b.
\(=\frac{2[(\sqrt{3}-(\sqrt{2}-1)]}{[(\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{3-(\sqrt{2}-1)^2}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{2\sqrt{2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{2}}=\frac{\sqrt{6}-2+\sqrt{2}}{2}\)
c.
\(=\frac{5(\sqrt[3]{2^2}-3\sqrt[3]{2}+3^2)}{(\sqrt[3]{2})^3+3^3}=\frac{5(\sqrt[3]{4}+3\sqrt[3]{2}+9)}{29}\)
a) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{1-2\sqrt{15}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
a) \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}+\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{8-2\sqrt{15}-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{1-2\sqrt{15}}\)
\(=\dfrac{\left(7\sqrt{5}-7\sqrt{3}+7\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{7\sqrt{5}+70\sqrt{3}-7\sqrt{3}-42\sqrt{5}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{-35\sqrt{5}+63\sqrt{3}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{35\sqrt{5}-63\sqrt{3}-7\sqrt{7}-14\sqrt{105}}{59}\)
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)
a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)
c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)
\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)
\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)
\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)
\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{4}\)
a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)
b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)
c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)
d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)
bạn hãy nhân ở mẫu với biểu thức tương ướng để tạo ra biểu thức liên hợp , là HĐT số 3 ạ