CMR:
a,\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}=8}\)
b,\(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
Những câu hỏi liên quan
Tính:Asqrt{20}-10sqrt{dfrac{1}{5}}+sqrt{left(sqrt{5}-1right)^2}B2sqrt{32}+5sqrt{8}-4sqrt{32}Csqrt{left(3-sqrt{2}^2right)}-sqrt{left(1-sqrt{2}right)^2}Dsqrt{left(5-1right)^2}+sqrt{left(sqrt{5}-3right)^2}Eleft(3+dfrac{5-sqrt{5}}{sqrt{5}-1}right)left(3-dfrac{5+sqrt{5}}{sqrt{5}-1}right)Fsqrt{6+2sqrt{5}}-sqrt{9-4sqrt{5}}Gdfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}Hdfrac{10}{sqrt{3}-1}-dfrac{55}{2sqrt{3}+1} help
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Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
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1)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
2)\(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
1) \(\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\\ =2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)=8=VP\)
=> ĐPCM
2) Bình phương 2 về, đpcm <=> \(2\sqrt{2}-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\) = \(2\left(\sqrt{2}-1\right)\)
VT = \(2\sqrt{2}-2\sqrt{2-1}=2\left(\sqrt{2}-1\right)\) = VP=> ĐPCM
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1) Ta có: \(\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6-2\sqrt{5}}}{2}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}{2}\)
=8
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Rút gọn các biểu thức sau:a) left(sqrt{8}-3sqrt{2}+sqrt{10}right)left(sqrt{2}-3sqrt{0,4}right) b) left(sqrt{28}-2sqrt{14}+sqrt{7}right).sqrt{7}+7sqrt{8}c) 2sqrt{left(sqrt{2}-3right)^2}+sqrt{2left(-3right)^2}-5sqrt{left(-1right)^4} d) left(frac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}-sqrt{5}}
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Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn các biểu thức sau :
a) left(sqrt{8}-3sqrt{2}+sqrt{10}right)sqrt{2}-sqrt{5}
b) 0,2sqrt{left(-10right)^2.3}+2sqrt{left(sqrt{3}-sqrt{5}right)^2}
c) left(dfrac{1}{2}sqrt{dfrac{1}{2}}-dfrac{3}{2}sqrt{2}+dfrac{4}{5}sqrt{200}right):dfrac{1}{8}
d) 2sqrt{left(sqrt{2}-3right)^2}+sqrt{2.left(-3right)^2}-5sqrt{left(-1right)^4}
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Rút gọn các biểu thức sau :
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
b) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
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giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Chứng minha,sqrt{9-4sqrt{5}}-sqrt{5}-2b,frac{sqrt{2}+1}{sqrt{2}-1}3+2sqrt{2}c,2sqrt{2}left(3-sqrt{2}right)+left(1+2sqrt{2}right)^2-2sqrt{6}9d,sqrt{frac{4}{left(2-sqrt{5}right)^2}}-sqrt{frac{4}{left(2+sqrt{5}right)^2}}8e,left(3+sqrt{5}right)left(sqrt{10}-sqrt{2}right)sqrt{3-sqrt{5}}8f,sqrt{sqrt{2}+1}-sqrt{sqrt{2}-1}sqrt{2left(sqrt{2}-1right)}
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Chứng minh
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
\(c,2\sqrt{2}\left(3-\sqrt{2}\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
\(e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(f,\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Ta có
:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|2-\sqrt{5}|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2=VP\left(đpcm\right)\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
Ta có:
\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)
\(=\frac{3+2\sqrt{2}}{2-1}\)
\(=3+2\sqrt{2}=VP\left(đpcm\right)\)
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c,Bạn xem lại đề
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)
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\(e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(VT=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)|1-\sqrt{5}|\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2\)
\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=\left(3+\sqrt{5}\right).2\left(3-\sqrt{5}\right)\)
\(=[3^2-\left(\sqrt{5}\right)^2].2\)
\(=4.2=8=VP\left(đpcm\right)\)
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Chứng minha,sqrt{9-4sqrt{5}}-sqrt{5}-2b,frac{sqrt{2}+1}{sqrt{2}-1}3+2sqrt{2}c,2sqrt{2}left(sqrt{3}-2right)+left(1+2sqrt{2}right)^2-2sqrt{6}9d,sqrt{frac{4}{left(2-sqrt{5}right)^2}}-sqrt{frac{4}{left(2+sqrt{5}right)^2}}8e,left(3+sqrt{5}right)left(10-sqrt{2}right)sqrt{3-sqrt{5}}8f,sqrt{sqrt{2}+1}-sqrt{sqrt{2}-1}sqrt{2left(sqrt{2}-1right)}
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Chứng minh
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
\(c,2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
\(e,\left(3+\sqrt{5}\right)\left(10-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(f,\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
rút gọna) left(-7sqrt{7}right)left(-2sqrt{8}right)b) -sqrt{33}.3sqrt{3}c) left(3sqrt{5}right).left(-10sqrt{3}right)d) dfrac{1}{2}sqrt{5}.left(-6sqrt{2}right)e) dfrac{2}{3}sqrt{7}.left(-dfrac{9}{16}sqrt{3}right)f) 15sqrt{6}:5sqrt{3}g) -25sqrt{12}:left(-5sqrt{6}right)h) 36sqrt{8}:12sqrt{2}i) 4sqrt{27}:left(-2sqrt{3}right)
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rút gọn
a) \(\left(-7\sqrt{7}\right)\left(-2\sqrt{8}\right)\)
b) \(-\sqrt{33}.3\sqrt{3}\)
c) \(\left(3\sqrt{5}\right).\left(-10\sqrt{3}\right)\)
d) \(\dfrac{1}{2}\sqrt{5}.\left(-6\sqrt{2}\right)\)
e) \(\dfrac{2}{3}\sqrt{7}.\left(-\dfrac{9}{16}\sqrt{3}\right)\)
f) \(15\sqrt{6}:5\sqrt{3}\)
g) \(-25\sqrt{12}:\left(-5\sqrt{6}\right)\)
h) \(36\sqrt{8}:12\sqrt{2}\)
i) \(4\sqrt{27}:\left(-2\sqrt{3}\right)\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
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rút gọn A)\(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}\)
B) \(\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}\)
C) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
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