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20 tháng 5 2019
a, phân tích vế trái ta được:
11+6\(\sqrt{2}\)=9+2.3.\(\sqrt{2}\)+2=(3+\(\sqrt{2}\))2\(\)=VP(dpcm)
b,phân tích vế trái ta được
\(\sqrt{11+6\sqrt{ }2}\)+\(\sqrt{11-6\sqrt{ }2}\)=|3+\(\sqrt{2}\)|+|3-\(\sqrt{2}\)|=6=VP(dpcm)
a,phân tích vế trái ta được
8-2\(\sqrt{7}\)=7-2\(\sqrt{7}\)+1=(\(\sqrt{7}\)-1)2
câu b sai đề nha
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2021
\(B=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121=-115\) là số nguyên (đpcm)
16 tháng 7 2021
b) Ta có: \(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121=-115
HS
Chứng minh:
a)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
b)\(8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\)
2
AT
14 tháng 6 2018
a/ \(11+6\sqrt{2}=9+2\cdot3\cdot\sqrt{2}+2=\left(3+\sqrt{2}\right)^2\left(đpcm\right)\)
b/ \(8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}\cdot1+1=\left(\sqrt{7}-1\right)^2\left(đpcm\right)\)
AV
14 tháng 6 2018
Ta có:a)11+6\(\sqrt{2}\) =\(9+2.3.\sqrt{2}+2\) =\(3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2\)=\(\left(3+\sqrt{2}\right)^2\)
b) \(8-2\sqrt{7}\) = \(7-2.1.\sqrt{7}+1\) =\(\left(\sqrt{7}\right)^2-2.1.\sqrt{7}+1^2\) =\(\left(\sqrt{7}-1\right)^2\)
27 tháng 7 2023
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
27 tháng 7 2020
Trả lời:
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(A=\sqrt{1}\)
\(A=1\)
\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=1\)
NN
20 tháng 9 2020
a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)
\(=\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)( đpcm )
28 tháng 7 2020
1: Chứng minh
a) Ta có: \(VT=11+6\sqrt{2}\)
\(=9+2\cdot3\cdot\sqrt{2}+2\)
\(=\left(3+\sqrt{2}\right)^2=VP\)(đpcm)
b) Ta có: \(VP=\left(\sqrt{7}-1\right)^2\)
\(=7-2\cdot\sqrt{7}\cdot1+1\)
\(=8-2\sqrt{7}=VT\)(đpcm)
c) Ta có: \(VT=\left(5-\sqrt{3}\right)^2\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=28-10\sqrt{3}=VP\)(đpcm)
d) Ta có: \(VP=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}-\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2=VT\)(đpcm)
TL
Trịnh Long
CTVVIP
28 tháng 7 2020
thêm dòng này nữa :33
⇔ 11 + \(6\sqrt{2}=11+6\sqrt{2}\left(đpcm\right)\)
Sửa đề CM: \(\sqrt{11+6\sqrt{2}}=3+\sqrt{2}\)
Ta có: \(\sqrt{11+6\sqrt{2}}=\sqrt{3^2+2.3\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=|3+\sqrt{2}|=3+\sqrt{2}\)
#)Sửa đề : \(\left(3+\sqrt{2}\right)^2\)
#)Giải :
\(11+6\sqrt{2}=9+2.3.\sqrt{2}+2=\left(3+\sqrt{2}\right)^2\left(đpcm\right)\)