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NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2021
\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
MN
1
9 tháng 6 2017
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-\sqrt{3}^2}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
TN
1
12 tháng 4 2019
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{20+4\sqrt{3}-10\sqrt{3}-6}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{4\left(5+\sqrt{3}\right)-2\sqrt{3}\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(4-2\sqrt{3}\right)\left(5+\sqrt{3}\right)}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\Rightarrow A=2\)
T
10 tháng 9 2023
a, \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\left|2-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot3+3^2}\)
\(=\sqrt{5}-2+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(=\sqrt{5}-2+\left|\sqrt{5}-3\right|\)
\(=\sqrt{5}-2+3-\sqrt{5}\)
\(=1\)
b, (ĐKXĐ: x ≥ 0; x ≠ 1)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{x-\sqrt{x}+3\sqrt{x}-3}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
#\(Toru\)
10 tháng 9 2023
a: \(=\sqrt{5}-2+3-\sqrt{5}=3-2=1\)
b:
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{x-5+\sqrt{x}-1+2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+\sqrt{x}-6+2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
24 tháng 11 2021
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
VP
29 tháng 5 2018
mình không viết lại đề nha
\(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2.\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(3+2\sqrt{3}+1\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(4+2\sqrt{3}\right).\left(14-6\sqrt{3}\right)}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{56-24\sqrt{3}+28\sqrt{3}-36}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{20+4\sqrt{3}}{5+\sqrt{3}}}\)
\(=\sqrt{\frac{\left(20+4\sqrt{3}\right).\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right).\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{100-20\sqrt{3}+20\sqrt{3}-12}{5^2-\sqrt{3}^2}}\)
\(=\sqrt{\frac{88}{25-3}}\)
\(=\sqrt{\frac{88}{22}}\)
\(=\sqrt{4}\)
\(=2\)
HỌC TỐT !!!
EM
3
12 tháng 6 2018
\(A=\frac{\left(\sqrt{3}+1\right)\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}=\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}=\frac{44\left(\sqrt{3}-1\right)}{22}=2\sqrt{3}-2\)
10 tháng 7 2017
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
\(\sqrt{\left(\sqrt{5}+3\right)^2}+\sqrt{14-6\sqrt{5}}\)\(=\left|\sqrt{5}+3\right|+\sqrt{9-2.3\sqrt{5}+5}\)
\(=\sqrt{5}+3+\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}+3+\left|3-\sqrt{5}\right|\)
\(=\sqrt{5}+3+3-\sqrt{5}=6\) ( do \(3-\sqrt{5}>0\))