Rút gọn biểu thức chứa căn
\(\sqrt{11-6\sqrt{2}}\)
K
Khách
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23 tháng 7 2023
1: =3+căn 2-3+căn 2
=2căn 2
2: =(căn 3-2)(căn 3+2)
=3-4=-1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 8 2021
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-\sqrt{2}-2\sqrt{2}+\sqrt{3}\right)=\dfrac{1}{\sqrt{2}}\left(2\sqrt{3}-3\sqrt{2}\right)\)
\(=\sqrt{6}-3\)
TK
1
21 tháng 12 2023
Sửa đề: \(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left|3-\sqrt{2}\right|+\left|\sqrt{2}-1\right|\)
\(=3-\sqrt{2}+\sqrt{2}-1\)
=3-1=2
22 tháng 7 2023
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
23 tháng 7 2023
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}\right)^2-11^2\)
\(=6-121\)
\(=-115\)
10 tháng 6 2016
\(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}-\sqrt{2}=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}-2}{\sqrt{2}}=\frac{\sqrt{11}+1-\left(\sqrt{11}-1\right)-2}{\sqrt{2}}=0\)
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.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
17 tháng 7 2023
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
23 tháng 7 2021
1.
Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)
\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)
\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)
\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)
24 tháng 7 2021
\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)
\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)
\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)
\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)
Trả lời:
\(\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2-6\sqrt{2}+9}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.3+3^2}\)
\(=\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left|\sqrt{2}-3\right|\)
\(=3-\sqrt{2}\) ( vì \(2< \sqrt{3}\Leftrightarrow2-\sqrt{3}< 0\))