\(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
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Tính:
\(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
Ta có: \(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{6}+2\right)}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(5+\sqrt{6}+\sqrt{6}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\dfrac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}{4}\)
\(=\dfrac{25-24}{4}=\dfrac{1}{4}\)
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pdfrac{2}{1-sqrt{2}}-dfrac{2}{1+sqrt{2}}
Qleft(dfrac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-dfrac{5}{sqrt{5}}right)cdotleft(sqrt{5}-sqrt{2}right)
Rdfrac{2}{7+4sqrt{3}}+dfrac{2}{7-4sqrt{3}}
Sdfrac{2}{sqrt{5}+1}-sqrt{dfrac{2}{3-sqrt{5}}}
Tdfrac{4}{1-sqrt{3}}-dfrac{sqrt{15}+sqrt{3}}{1+sqrt{5}}
Uleft(dfrac{1}{2-sqrt{5}}+dfrac{2}{sqrt{5}+sqrt{3}}right):dfrac{1}{sqrt{21-12sqrt{3}}}
Vdfrac{2}{sqrt{3}-1}-sqrt{dfrac{2}{6-3sqrt{3}}}
Wdfrac{5sqrt{3}}{sqrt{3-sqrt{5}}-sqrt{3}}-dfrac{5sqrt{3}}{sqrt{3-sqrt{5}}+sqr...
Đọc tiếp
\(p=\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(Q=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)\)
\(R=\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(S=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(T=\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(U=\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
\(V=\dfrac{2}{\sqrt{3}-1}-\sqrt{\dfrac{2}{6-3\sqrt{3}}}\)
\(W=\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
\(Y=\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
♡
\(\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(=\dfrac{2\left(1+\sqrt{2}\right)-2\left(1-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)
\(=\dfrac{2+2\sqrt{2}-2+2\sqrt{2}}{1-2}=-4\sqrt{2}\)
♡\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left[-\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-3\)
♡\(\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(=\dfrac{2\left(7-4\sqrt{3}\right)+2\left(7+4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{49-48}\)
= 28
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♡
\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{4}{6-2\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)-2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\sqrt{5}-2-2\sqrt{5}-2}{5-1}\)
= - 1
♡\(\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(=\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)}\)
\(=-2-2\sqrt{3}-\sqrt{3}=-2-3\sqrt{3}\)
♡\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\) (nhân [căn 2] vào cả tử và mẫu)
\(=\dfrac{2}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{2\left(5-\sqrt{5}\right)}{25-5}=\dfrac{5-\sqrt{5}}{10}\)
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\(A=\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}+2\sqrt{2}\\ B=\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(C=\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
\(D=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}+\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
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\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
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\(C=\dfrac{1}{2}\left(11+2\sqrt{30}\right)-\dfrac{\sqrt{30}}{2}-\dfrac{\sqrt{30}}{2}\\
=\dfrac{11}{2}+\sqrt{30}-\sqrt{30}\\
=\dfrac{11}{2}\)
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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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15 tháng 12 2021 lúc 21:58
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
Tính:
Adfrac{sqrt{10}-sqrt{2}}{sqrt{5}-1}-dfrac{2-sqrt{2}}{sqrt{2}-1}
Bleft(1+dfrac{2+sqrt{2}}{1+sqrt{2}}right)cdotleft(1-dfrac{2-sqrt{2}}{1-sqrt{2}}right)
Cleft(dfrac{sqrt{6}-sqrt{2}}{1-sqrt{3}}right):dfrac{1}{sqrt{2}-sqrt{3}}
Ddfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}:dfrac{1}{sqrt{6}}
Edfrac{3+2sqrt{3}}{sqrt{3}}+dfrac{2+sqrt{2}}{1+sqrt{2}}-dfrac{1}{2-sqrt{3}}
Đọc tiếp
Tính:
\(A=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}-\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(B=\left(1+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\right)\cdot\left(1-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
\(C=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{2}-\sqrt{3}}\)
\(D=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{6}}\)
\(E=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
a: \(A=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)
b: \(B=\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)=1-2=-1\)
c: \(B=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=-\sqrt{6}+2\)
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thực hiện phép tính
A=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
B=\(\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{2}}}\cdot\left(3+\sqrt{5}\right)\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=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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Bài : Thu gọn1) dfrac{3sqrt{5}-5sqrt{3}}{sqrt{15}-3}2) dfrac{sqrt{5+2sqrt{6}}}{sqrt{2}+sqrt{3}}3) dfrac{7+4sqrt{3}}{2+sqrt{3}}4) dfrac{16-6sqrt{7}}{sqrt{7}-3}5) dfrac{left(sqrt{3}-sqrt{2}right)^2+4sqrt{6}}{sqrt{3}+sqrt{2}}6) dfrac{left(sqrt{3}+2sqrt{5}right)^2-8sqrt{15}}{sqrt{6-2sqrt{10}}}
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Bài : Thu gọn
1) \(\dfrac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}\)
2) \(\dfrac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)
3) \(\dfrac{7+4\sqrt{3}}{2+\sqrt{3}}\)
4) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
5) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
6) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6-2\sqrt{10}}}\)
1.
\(\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}=\frac{3\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)
2.
\(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)
3.
\(\frac{7+4\sqrt{3}}{2+\sqrt{3}}=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
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4.
\(\frac{16-6\sqrt{7}}{\sqrt{7}-3}=\frac{3^2-2.3\sqrt{7}+7}{\sqrt{7}-3}=\frac{(\sqrt{7}-3)^2}{\sqrt{7}-3}=\sqrt{7}-3\)
5.
\(\frac{(\sqrt{3}-\sqrt{2})^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{2.3}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
6.
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{6-2\sqrt{10}}}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{6-2\sqrt{10}}}\)
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