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e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
a)
\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(2+2\sqrt{2}+1)}\)
\(=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)
b)
\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(3-2\sqrt{3}+1)(\sqrt{3}-1)}\)
\(=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)
c)
\((\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3=[(\sqrt[3]{4}+1-(\sqrt[3]{4}-1)][(\sqrt[3]{4}+1)^2+(\sqrt[3]{4}+1)(\sqrt[3]{4}-1)+(\sqrt[3]{4}-1)^2]\)
\(=2[\sqrt[3]{16}+1+2\sqrt[3]{4}+\sqrt[3]{16}-1+\sqrt[3]{16}+1-2\sqrt[3]{4}]\)
\(=2(3\sqrt[3]{16}+1)\)
d)
\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\)
\(=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)
e)
\(E=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Áp dụng công thức $(a+b)^3=a^3+b^3+3ab(a+b)$ ta có:
\(E^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}.E\)
\(E^3=40+3\sqrt[3]{20^2-(14\sqrt{2})^2}.E\)
\(E^3=40+3\sqrt[3]{8}.E=40+6E\)
\(\Leftrightarrow E^2(E-4)+4E(E-4)+10(E-4)=0\)
\(\Leftrightarrow (E-4)(E^2+4E+10)=0\)
Dễ thấy $E^2+4E+10=(E+2)^2+6\neq 0$ nên $E-4=0$ hay $E=4$
a, \(\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|=2+\sqrt{5}-2+\sqrt{5}=2\sqrt{5}\)
b, \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|-\left|1+\sqrt{3}\right|=\sqrt{3}-1-1-\sqrt{3}=2\)
c, \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
wq
\(=5-3=2\)