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tiểu long nữ
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lộc phạm
31 tháng 5 2018 lúc 21:35

Tu bieu thuc \(\Leftrightarrow\frac{3.\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{14\sqrt{7}}{7}+|\sqrt{7}-2|\)

                    \(\Leftrightarrow3\sqrt{7}+6-2\sqrt{7}+\sqrt{7}-2=2\sqrt{7}+4\)

hoaan
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Vũ Tiến Manh
9 tháng 10 2019 lúc 18:22

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) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)\(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)\(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)\(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)

nguyễn thị oanh
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Hoàng Lê Bảo Ngọc
13 tháng 8 2016 lúc 19:31

a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)

 

NT Ánh
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Trần Việt Linh
13 tháng 8 2016 lúc 21:09

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

shoppe pi pi pi pi
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Huong Ly Nguyen
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Hatsune Miku
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 ๖ۣۜFunny-Ngốkツ
30 tháng 8 2018 lúc 11:36

a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{1}{\sqrt{2}}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

An Tuệ
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thu  uyen
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Đinh Tuấn Việt
29 tháng 5 2015 lúc 22:17

\(=\frac{\sqrt{2}}{2}\)