(\(5\sqrt{3}\) + \(3\sqrt{5}\) ) : \(\sqrt{15}\)
Những câu hỏi liên quan
\(21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
Thu gọn B= \(21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{5}\)
Thu gọn A= \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
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TínhAleft(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}Bleft(3-sqrt{5}right)cdotsqrt{3+sqrt{5}}+left(3+sqrt{5}right)cdotsqrt{3-sqrt{5}}Csqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{ }}3}}Dsqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}Efrac{sqrt{15-10sqrt{2}}+sqrt{13+4sqrt{5}}-sqrt{11+2sqrt{10}}}{2sqrt{3+2sqrt{2}}+sqrt{9-4sqrt{2}}+sqrt{12+8sqrt{2}}}
Đọc tiếp
Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
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Thu gọn biểu thức:
\(B=21\cdot\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
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rút gọn các biểu thức sau:
a, sqrt{8-2sqrt{15}}-sqrt{8+2sqrt{15}}
b, dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}+dfrac{5-2sqrt{5}}{2sqrt{5}-4}
c, left(dfrac{1}{sqrt{5}-sqrt{2}}+dfrac{1}{sqrt{5}+sqrt{2}}right):dfrac{1}{left(sqrt{2}+1right)^2}
d,left(dfrac{2}{sqrt{3}-1}+dfrac{3}{sqrt{3}-2}+dfrac{15}{3-sqrt{3}}right).dfrac{1}{sqrt{3}+5}
Đọc tiếp
rút gọn các biểu thức sau:
a, \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
b, \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
c, \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
d,\(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)
a. \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
= \(\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}\)
= \(-2\sqrt{3}\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
= \(\dfrac{\left(\sqrt{15}-\sqrt{5}\right).\left(\sqrt{3}+1\right)}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(2\sqrt{5}+4\right)}{4}\)
=\(\dfrac{\sqrt{45}+\sqrt{15}-\sqrt{15}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).2\left(\sqrt{5}+2\right)}{4}\)
= \(\dfrac{3\sqrt{5}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(\sqrt{5}+2\right)}{2}\)
= \(\dfrac{2\sqrt{5}}{2}+\dfrac{5\sqrt{5}+10-10-4\sqrt{5}}{2}\)
= \(\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
= \(\dfrac{3\sqrt{5}}{2}\)
c. \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
= \(\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right).\left(\sqrt{5}+\sqrt{2}\right)}.\left(\sqrt{2}+1\right)^2\)
= \(\dfrac{2\sqrt{5}}{3}.\left(2+2\sqrt{2}+1\right)\)
= \(\dfrac{2\sqrt{5}}{3}.\left(3+2\sqrt{2}\right)\)
= \(\dfrac{6\sqrt{5}+4\sqrt{10}}{3}\)
d. \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-3\left(\sqrt{3}+2\right)+\dfrac{5\left(3+\sqrt{3}\right)}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(-2\sqrt{3}-5+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{-4\sqrt{3}-10+15+5\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{\sqrt{3}+5}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{1}{2}\)
Nếu đúng cho 1 like nhé!
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\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(\left(\sqrt{3}+1\right).\dfrac{\sqrt{3}-3}{2\sqrt{3}}\)
\(\dfrac{3\sqrt{18}-2\sqrt{8}}{\sqrt{50}}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{3\sqrt{5}}{2}\)
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\(\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}-3}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{2}\)
\(=\dfrac{1-3}{2}\)
\(=-1\)
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\(\dfrac{3\sqrt{18}-2\sqrt{8}}{\sqrt{50}}\)
\(=\dfrac{3\cdot3\sqrt{2}-2\cdot2\sqrt{2}}{5\sqrt{2}}\)
\(=\dfrac{9\sqrt{2}-4\sqrt{2}}{5\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}}{5\sqrt{2}}\)
\(=1\)
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6 tháng 7 2021 lúc 10:10
Thực hiện từng bước của phép tính:
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
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3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
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Rút gọn: \(M=\dfrac{8}{\sqrt{5}-\sqrt{3}}+\dfrac{7}{\sqrt{3}-2}+\dfrac{4}{\sqrt{2}-1}+\dfrac{3\sqrt{5}-\sqrt{15}}{\sqrt{15}}\)
\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)
\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)
\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)
\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)
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\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)
\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)
\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)
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tính:
P=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Q=\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
1) \(P=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2}\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2+\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+2\sqrt{60}+6\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+4\sqrt{15}+6\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(16+4\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{4\left(4+\sqrt{15}\right)\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{4\left(16-15\right)}\)
\(=\sqrt{4\cdot1}\)
\(=\sqrt{4}\)
\(=2\)
2) \(Q=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(3-\sqrt{5}\right)^2}\sqrt{3+\sqrt{5}}+\sqrt{\left(3+\sqrt{5}\right)^2}\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(3-\sqrt{5}\right)^2\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(9-6\sqrt{5}+5\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(14-6\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}+\sqrt{\left(9+6\sqrt{5}+5\right)\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{42+14\sqrt{5}-18\sqrt{5}-30}+\sqrt{42-14\sqrt{5}+18\sqrt{5}-30}\)
\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)
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\(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{15}}+\dfrac{4}{\sqrt{5}-3}+\dfrac{6}{\sqrt{3}}\)
Ta có: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{15}}+\dfrac{4}{\sqrt{5}-3}+\dfrac{6}{\sqrt{3}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{15}}-\dfrac{4\left(3+\sqrt{5}\right)}{4}+\sqrt{3}\cdot\dfrac{2\sqrt{3}}{\sqrt{3}}\)
\(=\sqrt{5}-\sqrt{3}-3-\sqrt{5}+2\sqrt{3}\)
\(=-3+\sqrt{3}\)
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