THỰC HIỆN PHÉP TÍNH
26) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
40)\(\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{\left|40\sqrt{2}+57\right|}\)
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D
0
VH
22 tháng 8 2020
a) \(E=\sqrt{\left|12\sqrt{5}-29\right|}-\sqrt{12\sqrt{5}+29}\)
\(\Leftrightarrow E^2=\left|12\sqrt{5}-29\right|-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=29-12\sqrt{5}-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=-24\sqrt{5}\)
\(\Leftrightarrow E=-2\sqrt{6\sqrt{5}}\)
b) Đặt \(F=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
\(\Leftrightarrow F^2=\left|40\sqrt{2}-57\right|-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=57-40\sqrt{2}-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=-80\sqrt{2}\)
\(\Leftrightarrow F=-4\sqrt{5\sqrt{2}}\)
5 tháng 8 2021
\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
DT
2
23 tháng 8 2018
\(L=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{\left|40\sqrt{2}-57\right|}\)
\(=\sqrt{40\sqrt{2}-57}-\sqrt{40\sqrt{2}-57}\)
\(=0\)
6 tháng 7 2021
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.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
6 tháng 7 2021
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
26, đặt bthuc là A suy ra A2=4+4+2\(\sqrt{16-\left(10+2\sqrt{5}\right)}\) suy ra A2=8+2(\(\sqrt{5}\) -1) suy ra A=\(\sqrt{6+2\sqrt{5}}\)=\(\sqrt{5}\)+1
40, tương tự
thanks p nhìu