\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}\) +\(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}_{ }+1}\)(\(\dfrac{\sqrt{3}}{2-\sqrt{6}}\)+\(\dfrac{\sqrt{3}}{2+\sqrt{6}}\))-\(\dfrac{1}{\sqrt{2}}\)
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Tính:1) dfrac{3}{1-sqrt{2}}+dfrac{sqrt{2}-1}{sqrt{2}+1}2) dfrac{sqrt{5}-1}{sqrt{5}+1}+dfrac{6}{1-sqrt{5}}3) dfrac{sqrt{2}+sqrt{3}}{2-sqrt{6}}+dfrac{sqrt{3}-sqrt{2}}{sqrt{6}+2}4) dfrac{3+sqrt{3}}{sqrt{3}}+dfrac{sqrt{6}-sqrt{3}}{1-sqrt{2}}5) dfrac{2-sqrt{2}}{1-sqrt{2}}+dfrac{sqrt{2}-sqrt{6}}{sqrt{3}-1}
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Tính:
1) \(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
2) \(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}\)
3) \(\dfrac{\sqrt{2}+\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)
4) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
5) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
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10 tháng 7 2023 lúc 11:10
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(\left(\dfrac{15}{3-\sqrt{3}}-\dfrac{2}{1-\sqrt{3}}+\dfrac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)
\(=\sqrt{6}+\sqrt{6}-1\)
\(=2\sqrt{6}-1\)
=======================
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)-3\left(\sqrt{2}-\sqrt{3}\right)}{-\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}\)
\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)
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1)\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
2)\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
3)\(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)
\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)
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Tính
\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=1\)
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b)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2\sqrt{6}}{6}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{6+3\sqrt{3}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}}{6\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{6+3\sqrt{3}}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}}{-2\sqrt{3}}\)
\(=\dfrac{3+\sqrt{3}}{-2\sqrt{3}}\)
\(=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}\)
\(=-\dfrac{3\sqrt{3}+3}{6}\)
\(=-\dfrac{3\left(\sqrt{3}+3\right)}{6}\)
\(=-\dfrac{\sqrt{3}+1}{2}\)
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\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
\(=\dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)\cdot\left(1-\sqrt{1+\dfrac{\sqrt{3}}{2}}\right)}{-\dfrac{\sqrt{3}}{2}}+\dfrac{\left(1-\dfrac{\sqrt{3}}{2}\right)\cdot\left(1+\sqrt{1-\dfrac{\sqrt{3}}{2}}\right)}{\dfrac{\sqrt{3}}{2}}\)
\(=\dfrac{1-\sqrt{1+\dfrac{\sqrt{3}}{2}}+\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3\left(1+\dfrac{\sqrt{3}}{2}\right)}}{2}}{-\dfrac{\sqrt{3}}{2}}+\dfrac{\left(1-\dfrac{\sqrt{3}}{2}\right)\cdot\left(1+\sqrt{1-\dfrac{\sqrt{3}}{2}}\right)\cdot2}{\sqrt{3}}\)
\(=\dfrac{1-\sqrt{1+\dfrac{\sqrt{3}}{2}}+\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3+\dfrac{3\sqrt{3}}{2}}}{2}}{-\dfrac{\sqrt{3}}{2}}+\dfrac{\left(2-\sqrt{3}\right)\cdot\left(1+\sqrt{1-\dfrac{\sqrt{3}}{2}}\right)}{\sqrt{3}}\)
\(=\dfrac{1-\sqrt{1+\dfrac{\sqrt{3}}{2}}+\dfrac{\sqrt{3}-\sqrt{3+\dfrac{3\sqrt{3}}{2}}}{2}}{\sqrt{3}}+\dfrac{2+2\sqrt{1-\dfrac{\sqrt{3}}{2}}-\sqrt{3}-\sqrt{3-\dfrac{3\sqrt{3}}{2}}}{\sqrt{3}}\)
\(=\dfrac{-\left(2-2\sqrt{1+\dfrac{\sqrt{3}}{2}}+\sqrt{3}-\sqrt{3+\dfrac{3\sqrt{3}}{2}}\right)+2\cdot2\sqrt{1-\dfrac{\sqrt{3}}{2}}-\sqrt{3}-\sqrt{3-\dfrac{3\sqrt{2}}{2}}}{\sqrt{3}}\)
\(=1\)
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Rút gọn:
a,
\(\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
\(a,\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{2}\right)}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{4-6}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{2}.\sqrt{3}}.\dfrac{4\sqrt{3}}{-2}-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{\sqrt{2}-\sqrt{3}-1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1+\left(\sqrt{2}-\sqrt{3}-1\right)\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}-1+2+\sqrt{6}-\sqrt{6}-3-\sqrt{2}-\sqrt{3}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{-2}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=-\dfrac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}\)
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chứng minh
\(\dfrac{3}{2}\)\(\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
rút gọn
D=\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\)\(-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
a)=\(\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{2\sqrt{6}}{3}-\dfrac{\sqrt{6}}{2} \)
=\(\dfrac{4\sqrt{6}}{6}-\dfrac{3\sqrt{6}}{6}=\dfrac{\sqrt[]{6}}{6}\)
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b)\(\dfrac{D}{\sqrt{3}}=\dfrac{\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1}{\sqrt{3}+1-1}\)
\(\dfrac{D}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
D=2
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5 câu:1) dfrac{sqrt{3}+sqrt{2}}{sqrt{6}+2}-dfrac{sqrt{3}-sqrt{2}}{sqrt{6}-2}2) dfrac{3}{sqrt{5}-sqrt{2}}-dfrac{2}{2-sqrt{2}}+dfrac{1}{sqrt{3}+sqrt{2}}3) dfrac{12}{sqrt{5}+1}-dfrac{4}{sqrt{5}+2}+dfrac{20}{3+sqrt{5}}4) dfrac{5}{3-sqrt{7}}-dfrac{2}{sqrt{2}+sqrt{3}}-dfrac{1}{sqrt{2}-1}5) dfrac{sqrt{12}-6}{sqrt{8}-sqrt{24}}-dfrac{3+sqrt{3}}{sqrt{3}}-dfrac{4}{sqrt{7}-1}
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5 câu:
1) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}+2}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}-2}\)
2) \(\dfrac{3}{\sqrt{5}-\sqrt{2}}-\dfrac{2}{2-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\)
4) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{2}{\sqrt{2}+\sqrt{3}}-\dfrac{1}{\sqrt{2}-1}\)
5) \(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{4}{\sqrt{7}-1}\)
1. dfrac{-2}{sqrt{3}-1}2. dfrac{5}{1-sqrt{6}}3. dfrac{2+sqrt{5}}{2-sqrt{5}}4. dfrac{1}{5+2sqrt{6}}5. dfrac{sqrt{5}+2}{sqrt{5}-2}6. dfrac{5sqrt{2}-2sqrt{5}}{sqrt{2}-sqrt{5}}7. dfrac{sqrt{20}-3sqrt{10}}{3-sqrt{2}}8. dfrac{6-2sqrt{5}}{3+sqrt{5}}9. dfrac{9+4sqrt{5}}{sqrt{5}+2}
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1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
Tính:1) dfrac{1}{3-2sqrt{2}}+dfrac{1}{2+sqrt{5}}2) dfrac{1}{3-2sqrt{2}}-dfrac{1}{3+2sqrt{2}}3) dfrac{1}{sqrt{5}-sqrt{7}}+dfrac{2}{1-sqrt{7}}4) dfrac{1}{5+2sqrt{6}}-dfrac{1}{5-2sqrt{6}}5) -dfrac{1}{sqrt{2}-sqrt{3}}-dfrac{3}{sqrt{18}+2sqrt{3}}
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Tính:
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{2+\sqrt{5}}\)
2) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
3) \(\dfrac{1}{\sqrt{5}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
4) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
5) \(-\dfrac{1}{\sqrt{2}-\sqrt{3}}\)\(-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
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Tính giá trị các biểu thức sau
1.dfrac{1}{sqrt{1}-sqrt{2}}-dfrac{1}{sqrt{2}-sqrt{3}}+dfrac{1}{sqrt{3}-sqrt{4}}-dfrac{1}{sqrt{4}-sqrt{5}}+dfrac{1}{sqrt{5}-sqrt{6}}-dfrac{1}{sqrt{6}-sqrt{7}}+dfrac{1}{sqrt{7}-sqrt{8}}-dfrac{1}{sqrt{8}-sqrt{9}}
2.dfrac{1}{2sqrt{1}+1sqrt{2}}+dfrac{1}{3sqrt{2}+2sqrt{3}}+dfrac{1}{4sqrt{3}+3sqrt{4}}+dfrac{1}{5sqrt{4}+4sqrt{5}}+dfrac{1}{6sqrt{5}+5sqrt{6}}+dfrac{1}{7sqrt{6}+6sqrt{7}}
giúp mk vs ạ
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Tính giá trị các biểu thức sau
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}\)
2.\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+\dfrac{1}{5\sqrt{4}+4\sqrt{5}}+\dfrac{1}{6\sqrt{5}+5\sqrt{6}}+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\)
giúp mk vs ạ
\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
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1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)
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