Thực hiện phép tình "
\(\sqrt{33-12\sqrt{6}}\)
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27 tháng 5 2021
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
27 tháng 5 2021
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).
21 tháng 7 2019
a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
= \(4\sqrt{5}\)
b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)
B
21 tháng 7 2019
\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=|3-\sqrt{6}|+|3-2\sqrt{6}|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
1
1
16 tháng 6 2021
\(D=\sqrt{9+6\sqrt{2}}-\sqrt{9-6\sqrt{2}}-\sqrt{21-12\sqrt{3}}\)
\(D=\sqrt{9+2.\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{9-2\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{21-2.2\sqrt{3}.3}\)
\(D=\sqrt{\left(\sqrt{6}\right)^2+2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(-\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}\)
\(D=\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{\left(3-2\sqrt{3}\right)^2}\)
\(D=\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}-2\sqrt{3}+3=3\)
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.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
NN
1
AH
Akai Haruma
Giáo viên
31 tháng 12 2020
Lời giải:
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)
\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)
KT
6 tháng 8 2018
\(\frac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\frac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)
\(=\frac{\sqrt{2}\left(2\sqrt{6}-\sqrt{3}\right)}{2\sqrt{6}-\sqrt{3}}+\frac{\sqrt{5}\left(2\sqrt{5}+1\right)}{\sqrt{3}\left(2\sqrt{5}+1\right)}\)
\(=\sqrt{2}+\frac{\sqrt{5}}{\sqrt{3}}\)
\(=\frac{\sqrt{6}+\sqrt{5}}{\sqrt{3}}\)
p/s: chúc bạn học tốt
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) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
\(=\sqrt{24+12\sqrt{6}+9}\)
\(=\sqrt{(2\sqrt{6})^2+2.2\sqrt{6}.3+3^2}\)
\(=\sqrt{(2\sqrt{6}+3)^2}\)
\(=|2\sqrt{6}+3|\)
\(=2\sqrt{6}+3\)