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Kkk
BL
10 tháng 7 2017
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
BL
10 tháng 7 2017
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
AH
Akai Haruma
Giáo viên
23 tháng 11 2018
Câu 1:
Có: \(8-4\sqrt{3}=8-2\sqrt{12}=6+2-2\sqrt{6.2}=(\sqrt{6}-\sqrt{2})^2\)
\(\Rightarrow \sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}\)
Do đó:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}\)
\(=\sqrt{(\sqrt{6})^2-(\sqrt{2})^2}=\sqrt{6-2}=2\)
AH
Akai Haruma
Giáo viên
23 tháng 11 2018
Câu 2:
\(16-5\sqrt{7}=\frac{32-10\sqrt{7}}{2}=\frac{32-2\sqrt{175}}{2}=\frac{25+7-2\sqrt{25.7}}{2}=\frac{(5-\sqrt{7})^2}{2}\)
\(\Rightarrow \sqrt{16-5\sqrt{7}}=\frac{5-\sqrt{7}}{\sqrt{2}}\)
Do đó:
\(\sqrt{16-5\sqrt{7}}(5\sqrt{2}+\sqrt{14})+\frac{6}{3+\sqrt{10}}=\frac{5-\sqrt{7}}{\sqrt{2}}.\sqrt{2}(5+\sqrt{7})+\frac{6(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})}\)
\(=(5-\sqrt{7})(5+\sqrt{7})+\frac{18-6\sqrt{10}}{3^2-10}=25-7+(-18+6\sqrt{10})\)
\(=6\sqrt{10}\)
9 tháng 7 2022
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
24 tháng 8 2017
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
24 tháng 8 2017
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
PA
14 tháng 7 2017
Bài 1:
\(M=\dfrac{9}{\sqrt{11}-\sqrt{2}}-\dfrac{\sqrt{22}-\sqrt{10}}{\sqrt{11}-\sqrt{5}}-\dfrac{22}{\sqrt{11}}\)
\(=\dfrac{9\left(\sqrt{11}+\sqrt{2}\right)}{11-2}-\dfrac{\sqrt{2}\left(\sqrt{11}-\sqrt{5}\right)\left(\sqrt{11}+\sqrt{5}\right)}{11-5}-\dfrac{2.\left(\sqrt{11}\right)^2}{\sqrt{11}}\)
\(=\sqrt{11}+\sqrt{2}-\sqrt{2}-2\sqrt{11}=-\sqrt{11}\)
\(M=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+\dfrac{a-b}{\sqrt{a}+\sqrt{b}}+\dfrac{2b}{\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\dfrac{2\left(\sqrt{b}\right)^2}{\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}+\sqrt{a}-\sqrt{b}+2\sqrt{b}=2\sqrt{a}\)
PA
14 tháng 7 2017
Bài 2:
a)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2}{\sqrt{x}+1}\) (*)
b)
Thay x = 0,25 vào (*), ta có:
\(M=\dfrac{2}{\sqrt{\dfrac{1}{4}}+1}=\dfrac{4}{3}\)
c)
\(M\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\ge1\)
\(\Leftrightarrow2\ge\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}\le1\)
\(\Leftrightarrow x\le1\)
mà x khác 1 và x > 0(theo ĐKXĐ)
=> 0 < x < 1 thì M \(\ge\) 1
18 tháng 6 2017
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)
C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)
C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)
C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)
C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)
18 tháng 6 2017
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)
D = \(-6\sqrt{5}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
8 tháng 8 2022
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
\(\dfrac{\sqrt{22}}{\sqrt{11}}\) + \(\sqrt{8}\) - \(\dfrac{7.\sqrt{2}}{3-\sqrt{2}}\)
= \(\sqrt{\dfrac{22}{11}}\) + 2\(\sqrt{2}\) - \(\dfrac{7.\sqrt{2}.(3+\sqrt{2})}{9-2}\)
= \(\sqrt{2}\) + 2\(\sqrt{2}\) - 3.\(\sqrt{2}\) - 2
= 3\(\sqrt{2}\)- 3\(\sqrt{2}\) - 2
= - 2