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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) \(B=\frac{3}{6+\sqrt{35}}-\frac{3}{6-\sqrt{35}}=\frac{3\left(6-\sqrt{35}\right)}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}-\frac{3\left(6+\sqrt{35}\right)}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}\)
\(=\frac{18-3\sqrt{35}-18-3\sqrt{35}}{36-35}=-6\sqrt{35}\)
3 tháng 2 2022
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
7 tháng 9 2020
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
7 tháng 9 2020
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)
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
\(=\frac{\sqrt{2}.\sqrt{6}-\sqrt{5}.\sqrt{6}}{\sqrt{6}}.\frac{\sqrt{5}.\sqrt{7}+\sqrt{2}.\sqrt{7}}{\sqrt{7}}\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
BT
29 tháng 5 2017
\(\sqrt{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=50\)
Vậy x = 50
b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
\(\Leftrightarrow\left(x+1\right)\sqrt{3}=2\sqrt{3}+3\sqrt{3}\)
\(\Leftrightarrow\left(x+1\right)\sqrt{3}=\left(2+3\right)\sqrt{3}\)
\(\Leftrightarrow x+1=5\)
\(\Leftrightarrow x=4\)
Vậy x = 4
29 tháng 5 2017
\(\sqrt{9\left(x-1\right)}=21\\9\left(x-1\right)=21^2\\x-1=49\\ x=48 \)\(\sqrt{3}x+\sqrt{3}=2\sqrt{3}+3\sqrt{3}\\ 0=\sqrt{3}\left(2+3-1-x\right)\\ 0=\sqrt{3}\left(4-x\right)\\ x=4\\ \)
12 tháng 6 2018
a) \(=\left(\sqrt{3}+2\right)^2\)
b)\(=\left(\sqrt{5}-\sqrt{2}\right)^2\)
c)\(=\left(\sqrt{5}+\sqrt{3}\right)^2\)
d)\(=\left(\sqrt{10}-\sqrt{2}\right)^2\)
e) \(=\left(\sqrt{7}+\sqrt{5}\right)^2\)
√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 \(\sqrt{7}-\sqrt{2}>0\)