\(\sqrt{12-2\sqrt{35}}\) giải giúp nhé
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25 tháng 9 2021
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
17 tháng 8 2021
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
3 tháng 2 2022
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
L
2
5 tháng 8 2020
\(\frac{\sqrt{12}-\sqrt{30}}{\sqrt{6}}\cdot\frac{\sqrt{35}+\sqrt{14}}{\sqrt{7}}\)
\(=\frac{2\sqrt{3}-\sqrt{30}}{\sqrt{6}}\cdot\frac{\sqrt{35}+\sqrt{14}}{\sqrt{7}}\)
\(=\frac{\left(2\sqrt{3}-\sqrt{30}\right)\cdot\left(\sqrt{35}+\sqrt{14}\right)}{\sqrt{6}\cdot\sqrt{7}}\)
\(=\frac{\left(2\sqrt{3}-\sqrt{30}\right)\cdot\left(\sqrt{35}+\sqrt{14}\right)}{\sqrt{42}}\)
\(=\frac{2\sqrt{3}\cdot\sqrt{35}+2\sqrt{3}\cdot\sqrt{14}-\sqrt{30}\cdot\sqrt{35}-\sqrt{30}\cdot\sqrt{14}}{\sqrt{42}}\)
\(=\frac{2\sqrt{105}+2\sqrt{42}-5\sqrt{42}-2\sqrt{105}}{\sqrt{42}}\)
\(=\frac{-3\sqrt{42}}{\sqrt{42}}=-3\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
5 tháng 8 2020
\(=\frac{\sqrt{2}.\sqrt{6}-\sqrt{5}.\sqrt{6}}{\sqrt{6}}.\frac{\sqrt{5}.\sqrt{7}+\sqrt{2}.\sqrt{7}}{\sqrt{7}}\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
NT
25 tháng 8 2021
\(\sqrt{12+2\sqrt{35}}=\sqrt{7+2\sqrt{7.5}+5}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}+\sqrt{5}\)
NW
28 tháng 6 2016
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
24 tháng 9 2016
Cho mình sửa đề xí ạ!
b) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
NN
28 tháng 8 2020
a) \(A=\left(1-\sqrt{18}+\sqrt{32}\right).\sqrt{3-2\sqrt{2}}\)
\(=\left(1-\sqrt{9.2}+\sqrt{16.2}\right).\sqrt{2-2\sqrt{2}+1}\)
\(=\left(1-\sqrt{9}.\sqrt{2}+\sqrt{16}.\sqrt{2}\right).\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(1-3\sqrt{2}+4\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
\(=\left(1+\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
Vì \(\sqrt{2}>1\)\(\Rightarrow\left|\sqrt{2}-1\right|>0\)
\(\Rightarrow A=\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)=\left(\sqrt{2}\right)^2-1=2-1=1\)
b) \(B=\frac{3}{6+\sqrt{35}}-\frac{3}{6-\sqrt{35}}=\frac{3\left(6-\sqrt{35}\right)}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}-\frac{3\left(6+\sqrt{35}\right)}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}\)
\(=\frac{18-3\sqrt{35}-18-3\sqrt{35}}{36-35}=-6\sqrt{35}\)
√12−2√35=?√4+√15=?
(3−√2)√11+6√2=?
(√5+√7)√12−2√35=?
√7−2√10−√7+2√10=?
√13−√160+√53+4√90
\(\sqrt{12-2\sqrt{35}}=\sqrt{12-2\sqrt{7.5}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.5}+\left(\sqrt{2}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|=\sqrt{7}-\sqrt{2}\)vì \(\sqrt{7}-\sqrt{2}>0\)