Tính:
\(\dfrac{\sqrt{2+\sqrt{3}}}{2}\cdot\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2}\right)\)
Những câu hỏi liên quan
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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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) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
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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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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\(A=\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}+2\sqrt{2}\\ B=\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(C=\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
\(D=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}+\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
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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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\(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
Quy đồng biểu thức ta được
=\(\dfrac{9+3\sqrt{6}-2\sqrt{6}-4+3\sqrt{6}-6+6+2\sqrt{6}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{5-2\sqrt{6}}{4}\)
=\(\dfrac{5+2\sqrt{6}}{1}\).\(\dfrac{5-2\sqrt{6}}{4}\)
=\(\dfrac{1}{4}\)
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Tính
a)\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{ }6}\)
b) \(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)
a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)
\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)
b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)
\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)
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P=\(\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a)tìm điều kiện để P có nghĩa
b)rút gọn P
c)tính giá trị của P với x=\(3+2\sqrt{2}\)
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
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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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