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8 tháng 7 2018
\(a.\left(4+\sqrt{7}\right)\left(\sqrt{14}-\sqrt{2}\right)\sqrt{4-\sqrt{7}}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)\sqrt{7-2\sqrt{7}+1}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)^2=2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)=2\left(16-7\right)=18\) \(b.\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}=\dfrac{4\sqrt{2}+\sqrt{14}}{6+\sqrt{7+2\sqrt{7}+1}}+\dfrac{4\sqrt{2}-\sqrt{14}}{6-\sqrt{7-2\sqrt{7}+1}}=\dfrac{4\sqrt{2}+\sqrt{14}}{7+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{7-\sqrt{7}}=\dfrac{\left(4\sqrt{2}+\sqrt{14}\right)\left(7-\sqrt{7}\right)+\left(4\sqrt{2}-\sqrt{14}\right)\left(7+\sqrt{7}\right)}{49-7}=\dfrac{28\sqrt{2}-4\sqrt{14}+7\sqrt{14}-7\sqrt{2}+28\sqrt{2}+4\sqrt{14}-7\sqrt{14}-7\sqrt{2}}{42}=\dfrac{42\sqrt{2}}{42}=\sqrt{2}\)
NN
13 tháng 5 2018
a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)
b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)
c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)
e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)
f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
23 tháng 6 2018
\(1.\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2+2-\sqrt{3}}{3-4}=-4\)\(2.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{8+6\sqrt{2}+6\sqrt{2}-8}{16-18}=\dfrac{-12\sqrt{2}}{2}-6\sqrt{2}\)\(3.\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.3+9}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\text{|}2\sqrt{2}-3\text{|}+\text{|}2\sqrt{2}+3\text{|}=4\sqrt{2}\)
\(4.\sqrt{29-4\sqrt{7}}-\sqrt{29+4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}.1+1}-\sqrt{28+2.2\sqrt{7}.1+1}=\sqrt{\left(2\sqrt{7}-1\right)^2}-\sqrt{\left(2\sqrt{7}+1\right)^2}=\text{|}2\sqrt{7}-1\text{|}-\text{|}2\sqrt{7}+1\text{|}=-2\)\(5.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\text{|}\sqrt{7}+1\text{|}-\text{|}\sqrt{7}-1\text{|}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}\)
23 tháng 6 2018
1)
\(\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}\)
\(=\dfrac{\left(\sqrt{3}+2\right)-\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}\)
\(=\dfrac{4}{\left(\sqrt{3}\right)^2-2^2}\)
\(=\dfrac{4}{3-4}=-4\)
22 tháng 8 2017
e) \(E=A-\sqrt{2}\)
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(A^2=8-2\sqrt{16-7}=8-6=2\)
\(A>0=>A=\sqrt{2}\)
\(E=A-\sqrt{2}=0\)
LD
26 tháng 9 2017
a)\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=\(\left(6\sqrt{10}+6\sqrt{2}-10\sqrt{2}-2\sqrt{10}\right)\sqrt{3+\sqrt{5}}\)
=\(\left(4\sqrt{10}-4\sqrt{2}\right)\sqrt{3+\sqrt{5}}=\left(4\sqrt{10}-4\sqrt{2}\right)\dfrac{\sqrt{5}+1}{2}\)
=\(\dfrac{20\sqrt{2}+4\sqrt{10}-4\sqrt{10}-4\sqrt{2}}{2}\)
=\(\dfrac{16\sqrt{2}}{2}=8\sqrt{2}\)
b)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
=\(\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=0\)
c)\(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)
=\(\dfrac{\sqrt{6}-1+\sqrt{6}+1}{\sqrt{2}}=2\sqrt{3}\)
d)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
=\(\dfrac{\sqrt{7}-1-\sqrt{7}-1+\sqrt{14}}{\sqrt{2}}=\sqrt{7}-1\)
e)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
=\(\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}=0\)
2 tháng 2 2022
3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)
5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)
\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)
Đặt \(S=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
\(\sqrt{2}S=\sqrt{2}.\sqrt{4-\sqrt{7}}-\sqrt{2}.\sqrt{4+\sqrt{7}}+\sqrt{2}.\sqrt{7}\)
\(\sqrt{2}S=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{2}.\sqrt{7}\)
\(\sqrt{2}S=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+\sqrt{2}.\sqrt{7}\)
\(\sqrt{2}S=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{2}.\sqrt{7}\)
\(\sqrt{2}S=2\sqrt{7}+\sqrt{2}.\sqrt{7}\)
\(\sqrt{2}S=\left(2+\sqrt{2}\right).\sqrt{7}\)
\(S=\frac{\left(2+\sqrt{2}\right).\sqrt{7}}{\sqrt{2}}\)
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