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12 tháng 8 2020
1)
BT = \(\sqrt{\left(5+\sqrt{3}\right)^2}-\sqrt{\left(5-\sqrt{3}\right)^2}\)
= \(\left(5+\sqrt{3}\right)-\left(5-\sqrt{3}\right)\)
= \(2\sqrt{3}\)
2:
BT = \(\frac{\sqrt{2}.\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\frac{5\left(\sqrt{6}-1\right)}{\left(1+\sqrt{6}\right)\left(\sqrt{6}-1\right)}\)
= \(\sqrt{6}-\frac{5\left(\sqrt{6}-1\right)}{5}\) = \(\sqrt{6}-\left(\sqrt{6}-1\right)\) =1
CH
0
S
9 tháng 7 2019
\(\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\left(vì:\sqrt{2}>\sqrt{1}=1\right)\)
\(\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{\left(2\right)^2+2.2\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\left(2+\sqrt{3}>0\right)\)
\(\sqrt{28-10\sqrt{3}}=\sqrt{25-10\sqrt{3}+3}=\sqrt{5^2-2.5\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\left(vì:5=\sqrt{25}>\sqrt{3}\right)\)
9 tháng 7 2019
Bạn giải thích cho mik chỗ \(\sqrt{\left(\sqrt{2}-1\right)^2}\)tại sao lại bằng Căn bậc 2 của 2 -1
5 tháng 6 2018
mình làm mẫu 2 bài nhé 2 bài kia bạn làm tương tự
1)a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{7}=\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}+\sqrt{7}=\sqrt{7}+\sqrt{3}+\sqrt{7}=2\sqrt{7}+\sqrt{3}\)
2)a) \(\sqrt{12-6\sqrt{3}}-\sqrt{3}=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{3}=3-\sqrt{3}-\sqrt{3}=3-2\sqrt{3}\)
b) \(\sqrt{7+2\sqrt{6}}-\sqrt{3}=\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{3}=1+\sqrt{6}-\sqrt{3}\)
AM
7 tháng 6 2019
\(\sqrt{11+6\sqrt{2}}-\sqrt{2}\text{=}\sqrt{9+6\sqrt{2}+4}-\sqrt{2}\)
=\(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{2}\text{=}3+\sqrt{2}-\sqrt{2}=3\)
HT
7 tháng 6 2019
a/ \(\sqrt{11+6\sqrt{2}}-\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{2}=3+\sqrt{2}-\sqrt{2}=3\)
b/ \(\sqrt{28-10\sqrt{3}}+5=\sqrt{25-10\sqrt{3}+3}+5\)
\(=\sqrt{\left(5-\sqrt{3}\right)^2}+5=5-\sqrt{3}+5=25-\sqrt{3}\)
c/ \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
d/ \(3\sqrt{5}-\sqrt{6-2\sqrt{5}}=3\sqrt{5}-\sqrt{5-2\sqrt{5}+1}\)\
\(=3\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=3\sqrt{5}-\sqrt{5}+1=2\sqrt{5}+1\)
27 tháng 7 2020
a, \(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\sqrt{2}\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}=2+\sqrt{2}-\sqrt{6}\)
b, \(=\sqrt{300.0,04}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c, \(=\sqrt{196}-2\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
d, \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)
27 tháng 7 2020
Bài 1: Rút gọn
a) Ta có: \(\left(\sqrt{3}-\sqrt{2}+1\right)\cdot\left(\sqrt{3}-1\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)-\sqrt{2}\cdot\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}\)
\(=2-\sqrt{2}-\sqrt{6}\)
b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=0.2\cdot\sqrt{\left(-10\right)^2}\cdot\sqrt{3}+2\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=0.2\cdot10\cdot\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
c) Ta có: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\sqrt{196}-2\cdot\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-\sqrt{392}+7+\sqrt{392}\)
=21
d) Ta có: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)
\(=16\sqrt{5}\)
17 tháng 7 2019
\(\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
\(=\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=\left(\frac{5\sqrt{3}}{\sqrt{3}-1}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{5-3\sqrt{3}}:\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{\left(5-3\sqrt{3}\right)\left(5+3\sqrt{3}\right)}\)
\(=-\frac{5\sqrt{3}-8}{16-10\sqrt{3}}\)
= -(-1/2) = 1/2
1231+23456=
\(\sqrt{28+10\sqrt{3}}\)=\(\sqrt{\left(\sqrt{3}\right)^2+2.5.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{25}.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+5\right)^2}\)
=\(|\sqrt{3}+5|\)
=\(\sqrt{3}+5\)(vì \(\sqrt{3}+5\)\(\ge0\))