(\(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\)):\(\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
Những câu hỏi liên quan
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{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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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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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) \(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{4}{\sqrt{7}-1}\)
Rút gọn: ( 2,5 Điểm )A dfrac{sqrt{6+2sqrt{5}}}{sqrt{5}+1}+ dfrac{sqrt{5-2sqrt{6}}}{sqrt{3}-sqrt{2}}B dfrac{3}{sqrt{5}-2}+ dfrac{4}{sqrt{6}+sqrt{2}}+ dfrac{1}{sqrt{6}+sqrt{5}}C dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{99}+sqrt{100}}D dfrac{1}{2-sqrt{3}}+sqrt{7-4sqrt{3}}E sqrt{dfrac{3sqrt{3}-4}{2sqrt{3}+1}}-sqrt{dfrac{sqrt{3}+4}{5-2sqrt{3}}}F dfrac{1}{2+sqrt{3}}+dfrac{sqrt{2}}{sqrt{6}}-dfrac{2}{3+sqrt{3}}
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Rút gọn: ( 2,5 Điểm )
A= \(\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)+ \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
B= \(\dfrac{3}{\sqrt{5}-2}\)+ \(\dfrac{4}{\sqrt{6}+\sqrt{2}}\)+ \(\dfrac{1}{\sqrt{6}+\sqrt{5}}\)
C = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
D= \(\dfrac{1}{2-\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
E = \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
F = \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
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Rút gọn:a)dfrac{5sqrt{2}-2sqrt{5}}{sqrt{5}-sqrt{2}}+dfrac{6}{2-sqrt{10}}b)dfrac{6}{sqrt{5}-1}+dfrac{7}{1-sqrt{3}}-dfrac{2}{sqrt{3}-sqrt{5}}c)left(dfrac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right)divdfrac{1}{sqrt{7}-sqrt{5}}d)sqrt{2}+dfrac{1}{sqrt{5+2sqrt{6}}}+dfrac{2}{sqrt{8+2sqrt{15}}}e)left(dfrac{15}{sqrt{6}+1}+dfrac{4}{sqrt{6}-2}-dfrac{12}{3-sqrt{6}}right)timesleft(sqrt{6}+11right)Lm nhanh giúp mk nhé, mk đang cần gấp!
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Rút gọn:
a)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b)\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
c)\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d)\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
e)\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\times\left(\sqrt{6}+11\right)\)
Lm nhanh giúp mk nhé, mk đang cần gấp!
Bạn chia nhỏ ra để nhận được câu tl sớm nhất nhé!Bạn đặt câu hỏi free mà để dày cộp như này khum ai dám làm =(((
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\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{2\sqrt{2}}{2+\sqrt{2}}-\dfrac{4\sqrt{2}+4}{2+\sqrt{2}}\)
\(=\sqrt{3}+\sqrt{3}+\dfrac{-2\sqrt{2}-4}{2+\sqrt{2}}\)
\(=2\sqrt{3}+\dfrac{-2\left(2+\sqrt{2}\right)}{2+\sqrt{2}}\)
\(=2\sqrt{3}-2\)
\(------\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\dfrac{5\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{5\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)
\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)
\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)
\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)
\(=\dfrac{19\sqrt{5}-29}{4}\)
#Ayumu
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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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12 tháng 10 2021 lúc 19:44
* Thực hiện phép tính
a, \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
b. \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
c. \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
=-5+2
=-3
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dfrac{2sqrt{3}-3sqrt{2}}{sqrt{6}}-dfrac{2}{1-sqrt{3}}dfrac{4}{sqrt{6}+sqrt{2}}-dfrac{sqrt{54}+sqrt{2}}{sqrt{3}+1}dfrac{5+2sqrt{5}}{sqrt{5}}-dfrac{20}{5+sqrt{5}}-sqrt{20}Bài 2sqrt{25x^2-10x+1}sqrt{4x^2+8x+4}sqrt{x^2-3}+1xsqrt{7-2x}sqrt{x^2+7}sqrt{9x-27}+dfrac{1}{2}sqrt{4x-12}-9sqrt{dfrac{x-3}{9}}2
Đọc tiếp
\(\dfrac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}-\dfrac{2}{1-\sqrt{3}}\)
\(\dfrac{4}{\sqrt{6}+\sqrt{2}}-\dfrac{\sqrt{54}+\sqrt{2}}{\sqrt{3}+1}\)
\(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
Bài 2
\(\sqrt{25x^2-10x+1}=\sqrt{4x^2+8x+4}\)
\(\sqrt{x^2-3}+1=x\)
\(\sqrt{7-2x}=\sqrt{x^2+7}\)
\(\sqrt{9x-27}+\dfrac{1}{2}\sqrt{4x-12}-9\sqrt{\dfrac{x-3}{9}}=2\)
\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)
\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
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Bài 1:
d: Ta có: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
\(=\sqrt{5}+2-5+\sqrt{5}-2\sqrt{5}\)
=-3
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