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a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
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$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
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C làm giống câu a, nhé.
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\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
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\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)
\(=\sqrt{2\cdot\left(4+\sqrt{7}\right)}+\sqrt{2\cdot\left(4-\sqrt{7}\right)}\)
\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)
\(=\sqrt{7}+1+\sqrt{7}-1\)
\(=2\sqrt{7}\)
b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)
\(=-\dfrac{2}{\sqrt{2}}\)
\(=-\sqrt{2}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
a) Ta có: \(4\sqrt{28}+3\sqrt{63}-3\sqrt{112}-2\sqrt{175}\)
\(=8\sqrt{7}+9\sqrt{7}-12\sqrt{7}-10\sqrt{7}\)
\(=-5\sqrt{7}\)
b) Ta có: \(\sqrt{5}\left(\sqrt{5}-3\sqrt{20}+2\sqrt{80}\right)\)
\(=\sqrt{5}\left(\sqrt{5}-6\sqrt{5}+8\sqrt{5}\right)\)
\(=\sqrt{5}\cdot3\sqrt{5}=15\)
c) Ta có: \(\left(\sqrt{\dfrac{16}{3}}-\sqrt{\dfrac{25}{3}}\right)\cdot\sqrt{3}\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\sqrt{3}\)
=-1
e) Ta có: \(\left(\sqrt{\dfrac{32}{3}}-\sqrt{54}+\sqrt{\dfrac{50}{3}}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{4\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-3\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\dfrac{9\sqrt{12}}{\sqrt{3}}-18\)
\(=0\)
f) Ta có: \(\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3\sqrt{2}+2\sqrt{3}-2\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
\(A=4-7+6=3\)
\(B=\sqrt{4.7^2}-2\sqrt{25.7^2}+\sqrt{9.7^2}=2.7-2.5.7+3.7=-35\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}.4=4\left(\sqrt{3}-1\right)=4\sqrt{3}-4\)
\(D=\sqrt{\left(2+\sqrt{3}\right)^2}.\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\)