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MN
18 tháng 8 2021
a, \(14\sqrt{\frac{1}{7}}-\frac{3}{2}\sqrt{28}\)\(+20\sqrt{0,63}\)
\(=2\sqrt{7}-3\sqrt{7}+6\sqrt{7}\)
\(=5\sqrt{7}\)
b, \(\sqrt{\frac{a}{2}}+\frac{4}{5}\sqrt{8\text{a}}-\sqrt{\frac{2\text{a}}{9}}v\text{ới}a\ge0\)
\(=\sqrt{\frac{2\text{a}}{4\text{ }}}+\frac{8}{5}\sqrt{2\text{a}}-\frac{1}{3}\sqrt{2\text{a}}\)
\(=\frac{1}{2}\sqrt{2\text{a}}+\frac{8}{5}\sqrt{2\text{a}}-\frac{1}{3}\sqrt{2\text{a}}\)
\(=\frac{53}{30}\sqrt{2\text{a}}\)
HP
2
19 tháng 6 2017
\(A=\sqrt{9.7}-2\sqrt{25.7}+\sqrt{9.7.4}-\frac{1}{7}\sqrt{4.7}\)
\(=3\sqrt{7}-10\sqrt{7}+6\sqrt{7}-\frac{2}{7}\sqrt{7}\)
\(=\frac{-9}{7}\sqrt{7}\)
Nếu đúng tk nhé
19 tháng 6 2017
a = \(\sqrt{63}-2\sqrt{175}+\sqrt{252}-\frac{1}{7}\sqrt{28}\)
= \(\sqrt{\frac{4}{7}}\left(1,5-5+3-1\right)\)
= \(-1,5\sqrt{\frac{4}{7}}\)
VT
9 tháng 10 2019
a)= \(\left(3+\sqrt{5}\right)\left(\sqrt{\left(3-\sqrt{5}\right)^2}\right)\)=\(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)
b)= \(\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{\sqrt{2^2.7}}{2}-2\)=\(\frac{2\left(3-\sqrt{7}\right)}{9-7}+\sqrt{7}-2\)=1
c) =\(\frac{3}{3\left(\sqrt{7}-2\right)}-\frac{3}{3\left(\sqrt{7}+2\right)}\)=\(\frac{1}{\sqrt{7}-2}-\frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2-\left(\sqrt{7}-2\right)}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)}\)=\(\frac{4}{7-4}=\frac{4}{3}\)
d) =\(\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)^{ }\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{\left(88-44\sqrt{3}\right)}{25-3}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{22\left(4-2\sqrt{3}\right)}{22}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)\)=3-1 = 2
e) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)= \(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)= \(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)= \(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)
12 tháng 7 2016
a) ĐKXĐ : \(0\le a\ne1\)
\(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
b) ĐKXĐ : \(b\ne0,a\ne-\sqrt{b}\)
\(\frac{a-\sqrt{b}}{\sqrt{b}}:\frac{\sqrt{b}}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\sqrt{b}}.\frac{a+\sqrt{b}}{\sqrt{b}}=\frac{a^2-b}{b}=\frac{a^2}{b}-1\)
c) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=\sqrt{5}\left(2-5-4+11\right)\)\(=4\sqrt{5}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=7\left(2-2\sqrt{2}+1\right)+14\sqrt{2}=7\left(2-2\sqrt{2}+1+2\sqrt{2}\right)=7.3=21\)
e) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
25 tháng 6 2019
Bn có thể qua hoc.24h.vn hỏi nha , ở đó có nhiều người biết đó
~ Hok tốt ~
#Gumball
9 tháng 7 2019
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
\(A=\frac{2}{5+\sqrt{7}}+\frac{\sqrt{28}}{2}-2\)
\(A=\frac{2.\left(5-\sqrt{7}\right)}{25-7}+\frac{2\sqrt{7}}{2}-2\)
\(A=\frac{2.\left(5-\sqrt{7}\right)}{18}+\sqrt{7}-2\)
\(A=\frac{5-\sqrt{7}}{9}+\sqrt{7}-2\)
\(A=\frac{5-\sqrt{7}+9\sqrt{7}-18}{9}\)
\(A=\frac{-13+8\sqrt{7}}{9}\)
Vậy \(A=\frac{-13+8\sqrt{7}}{9}\)
\(A=\frac{2}{5+\sqrt{7}}+\frac{\sqrt{28}}{2}-2\)
\(=\frac{2\left(5-\sqrt{7}\right)}{25-7}+\frac{2\sqrt{7}}{2}-2\)
\(=\frac{2\left(5-\sqrt{7}\right)}{18}+\sqrt{7}-2\)
\(=\frac{2\left(5-\sqrt{7}\right)}{2.9}+\sqrt{7}-2=\frac{5-\sqrt{7}}{9}+\sqrt{7}-2\)