Tính
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
b) \(\sqrt{9-4\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
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AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
27 tháng 7 2021
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
PN
19 tháng 8 2020
\(a,\frac{2}{\sqrt{2}-1}-\frac{2}{\sqrt{2}+1}=\frac{2\left(\sqrt{2}+1\right)-2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2\sqrt{2}+2-2\sqrt{2}+2}{\sqrt{2}^2-1^2}=\frac{4}{2-1}=4\)
\(b,\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{4+2.2.\sqrt{2}+2}+\sqrt{4-2.2.\sqrt{2}+2}\)
\(=\sqrt{2^2+2.2.\sqrt{2}+\sqrt{2}^2}+\sqrt{2^2-2.2.\sqrt{2}+\sqrt{2}^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=|2+\sqrt{2}|+|2-\sqrt{2}|=2+2=4\)
\(c,\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+2.2.\sqrt{5}+5}+\sqrt{4-2.2.\sqrt{5}+5}\)
\(=\sqrt{2^2+2.2.\sqrt{5}+\sqrt{5}^2}+\sqrt{2^2-2.2.\sqrt{5}+\sqrt{5}^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|+|2-\sqrt{5}|=2+\sqrt{5}+\sqrt{5}-2=2\sqrt{5}\)
câu d bạn cứ nhân bình thường
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
7.
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
6 tháng 7 2021
a) Ta có: \(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) Ta có: \(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{2}+2\sqrt{2}+1}\)
\(=\sqrt{14+32\sqrt{2}}\)
c) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{3}-1}\)
\(=\sqrt{5+2\sqrt{5}-2\sqrt{3}}\)
Y
14 tháng 8 2023
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
24 tháng 9 2021
\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)
24 tháng 9 2021
a) \(=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}\)
b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)
c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)
6 tháng 6 2018
a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
b,c tương tự
a) \(\sqrt{3-2\sqrt{2}}\)+\(\sqrt{3+2\sqrt{2}}\)
= \(\sqrt{2-2\sqrt{2}+1}\)+ \(\sqrt{2+2\sqrt{2}+1}\)
= \(\sqrt{\left(\sqrt{2}-1\right)^2}\)+ \(\sqrt{\left(\sqrt{2}+1\right)^2}\)
= \(\sqrt{2}\)-1+\(\sqrt{2}\)+1
=2\(\sqrt{2}\)
b) \(\sqrt{9-4\sqrt{5}}\)+ \(\sqrt{6+2\sqrt{5}}\)
= \(\sqrt{4-4\sqrt{5}+5}\)+\(\sqrt{4+2\sqrt{5}+2}\)
= \(\sqrt{\left(2-\sqrt{5}\right)^2}\)+\(\sqrt{\left(2+\sqrt{2}\right)^2}\)
= 2-\(\sqrt{5}\)+2+\(\sqrt{2}\)
= 4-\(\sqrt{5}\)+\(\sqrt{2}\)