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Trần Nguyễn
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Nguyễn Lê Phước Thịnh
24 tháng 8 2021 lúc 15:17

a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)

=3

Nguyễn Nhã Thanh
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Ly Ly
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Nguyễn Hoàng Minh
12 tháng 10 2021 lúc 19:38

\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)

\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)

\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)

Nguyễn Nhã Thanh
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 ๖ۣۜFunny-Ngốkツ
11 tháng 6 2018 lúc 21:00

\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}\right)^2-2^2}-\frac{5\left(4-\sqrt{7}\right)}{4^2-\left(\sqrt{7}\right)^2}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{8}\)

\(=\frac{12\left(\sqrt{7}-5\right)}{24}-\frac{6\left(6+2\sqrt{7}\right)}{24}+\frac{8\left(6\sqrt{7}+12\right)}{24}-\frac{3\left(20-5\sqrt{7}\right)}{24}\)

\(=\frac{12\sqrt{7}-60-36-12\sqrt{7}+48\sqrt{7}+96-60+15\sqrt{7}}{24}\)

\(=\frac{-60+63\sqrt{7}}{24}\)

Nguyễn Nhã Thanh
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Phạm Hà Linh
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HT.Phong (9A5)
8 tháng 9 2023 lúc 5:59

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

Nguyễn Nhã Thanh
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Ánh Hoàng
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Hồng Phúc
3 tháng 9 2021 lúc 13:47

b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)

\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)

\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)

Hồng Phúc
3 tháng 9 2021 lúc 13:46

a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)

\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)

Hồng Phúc
3 tháng 9 2021 lúc 13:48

c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)

\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)

\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)

\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)