(2\sqrt(3)-3\sqrt(2))/(\sqrt(3)-\sqrt(2))+(5)/(1+\sqrt(6))-6\sqrt((1)/(6))
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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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1, dfrac{6-sqrt{6}}{sqrt{6}-1}+dfrac{6+sqrt{6}}{sqrt{6}}
2, dfrac{6-6sqrt{3}}{1-sqrt{3}}+dfrac{3sqrt{3}+3}{sqrt{3}+1}
3, dfrac{3+sqrt{3}}{sqrt{3}}+dfrac{sqrt{6}-sqrt{3}}{1-sqrt{2}}
4, dfrac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+dfrac{6+2sqrt{6}}{sqrt{3}+sqrt{2}}
5, left(dfrac{3sqrt{125}}{15}-dfrac{10-4sqrt{5}}{sqrt{5}-2}right)cdotdfrac{1}{sqrt{5}}
Đọc tiếp
1, \(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6+\sqrt{6}}{\sqrt{6}}\)
2, \(\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
3, \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
4, \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
5, \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\cdot\dfrac{1}{\sqrt{5}}\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
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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+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. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
Câu 1: Thực hiện phép tính:
a. \(\sqrt{3}\left(2\sqrt{6}-\sqrt{3}\right)-6\sqrt{2}\)
b. \(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\)
c. \(\sqrt{\left(1-\sqrt{3}\right)^2}-3\sqrt{\dfrac{1}{3}}\)
d. \(\dfrac{6}{\sqrt{6}}-\dfrac{5}{\sqrt{6}-1}\)
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
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1. Tính
a) \(\sqrt[3]{(\sqrt{2}+3)(11+6\sqrt{2})}\sqrt[3]{(\sqrt{2}+-3)(11-6\sqrt{2})}\)
b) (\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}-\sqrt[3]{2})\)
c)\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
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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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:a) sqrt{dfrac{1}{8}}.sqrt{2}.sqrt{125}.sqrt{dfrac{1}{5}}b)sqrt{sqrt{2}-1}.sqrt{sqrt{2}+1}c) sqrt{11-6sqrt{2}}.sqrt{11+6sqrt{2}}d) sqrt{12-6sqrt{3}}.sqrt{dfrac{1}{3-sqrt{3}}}e) dfrac{sqrt{15}-sqrt{6}}{sqrt{35}-sqrt{14}}f) dfrac{2sqrt{15}-2sqrt{10}+sqrt{6}-3}{2sqrt{5}-2sqrt{10}-sqrt{3}+sqrt{6}}g) left(dfrac{1}{5-2sqrt{6}}+dfrac{2}{5+2sqrt{6}}right)left(15+2sqrt{6}right)
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tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
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b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
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c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)
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Xem thêm câu trả lời
GIÚP EM ĐI ẠTÍNH:frac{3-sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3-sqrt{3+sqrt{6+sqrt{3}}}}+frac{2+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}{3+sqrt{6+sqrt{3+sqrt{6+sqrt{3}}}}}frac{1+frac{sqrt{3}}{2}}{1+sqrt{1+frac{sqrt{3}}{2}}}+frac{1-frac{sqrt{3}}{2}}{1-sqrt{1-frac{sqrt{3}}{2}}}frac{1}{sqrt{frac{5}{13}}+sqrt{frac{5}{7}}+1}+frac{1}{sqrt{frac{7}{5}}+sqrt{frac{7}{13}}+1}+frac{1}{sqrt{1frac{6}{7}}+1+sqrt{2frac{3}{5}}}RÚT GỌNsqrt{left(x-1right)^2}-x với x lớn hơn 1GIẢI PHƯƠNG TRÌNHsqrt{x^2-3x+2}+sqrt{x^2-4x+3}0
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GIÚP EM ĐI Ạ
TÍNH:
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(\frac{1}{\sqrt{\frac{5}{13}}+\sqrt{\frac{5}{7}}+1}+\frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{7}{13}}+1}+\frac{1}{\sqrt{1\frac{6}{7}}+1+\sqrt{2\frac{3}{5}}}\)
RÚT GỌN
\(\sqrt{\left(x-1\right)^2}-x\) với x lớn hơn 1
GIẢI PHƯƠNG TRÌNH
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
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1) frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)2) 0.1sqrt{left(-3right)^2}cdotleft[6sqrt{left(-frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^23) left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)4) left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}5) sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
Đọc tiếp
1) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
2) \(0.1\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(-\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
3) \(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
4) \(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
5) \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
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\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
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