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A=\(\sqrt{27}-6\sqrt{\dfrac{1}{3}+\dfrac{\sqrt{3}-3}{\sqrt{3}}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 8 2021
\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)
\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)
\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn
19 tháng 10 2021
c: \(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}\)
\(=\dfrac{23}{15}\sqrt{3}\)
3 tháng 7 2023
a: \(=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
b: \(=\dfrac{\sqrt{10}\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\left(\sqrt{11}+\sqrt{7}\right)}=\sqrt{\dfrac{10}{2}}=\sqrt{5}\)
c: \(=\dfrac{\sqrt{6}\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{7}-\sqrt{6}\right)}=\sqrt{\dfrac{6}{3}}=\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 7 2023
1) \(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\sqrt{5}+3\sqrt{9\cdot3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\sqrt{5}+3\cdot3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9\cdot\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{9}{1}=9\)
2) \(\dfrac{\sqrt{110}+\sqrt{70}}{\sqrt{22}+\sqrt{14}}\)
\(=\dfrac{\sqrt{10}\cdot\sqrt{11}+\sqrt{10}\cdot\sqrt{7}}{\sqrt{2}\cdot\sqrt{11}+\sqrt{2}\cdot\sqrt{7}}\)
\(=\dfrac{\sqrt{10}\cdot\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\cdot\left(\sqrt{11}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{10}}{\sqrt{2}}=\sqrt{\dfrac{10}{2}}\)
\(=\sqrt{5}\)
3) \(\dfrac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{7}-\sqrt{6}\cdot\sqrt{6}}{\sqrt{3}\cdot\sqrt{7}-\sqrt{3}\cdot\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{3}\cdot\left(\sqrt{7}-\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{3}}=\sqrt{\dfrac{6}{3}}\)
\(=\sqrt{2}\)
19 tháng 1 2022
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
19 tháng 1 2022
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
TH
Thầy Hùng Olm
Manager
VIP
5 tháng 7 2023
a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)
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