trục căn thức ở mẫu
18√14-60/2(3√7-5√2)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
24 tháng 10 2021
\(\dfrac{-2}{\sqrt{3}-1}=\dfrac{-2\left(\sqrt{3}+1\right)}{2}=-\sqrt{3}-1\\ \dfrac{\sqrt{5}}{\sqrt{7}-3}=\dfrac{-\sqrt{5}\left(\sqrt{7}+3\right)}{2}\\ \dfrac{3\sqrt{3}-2}{1-2\sqrt{3}}=\dfrac{\left(3\sqrt{3}-2\right)\left(1+2\sqrt{3}\right)}{-11}=\dfrac{\sqrt{3}-16}{11}\\ \dfrac{14}{\sqrt{10}+\sqrt{3}}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{7}=2\sqrt{10}-2\sqrt{3}\)
XT
1
22 tháng 7 2018
a) \(\dfrac{14}{2\sqrt{3}-\sqrt{5}}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}\right)^2-\sqrt{5^2}}=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{12-5}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{7}=2\left(2\sqrt{3}+\sqrt{5}\right)\)
\(=4\sqrt{3}+2\sqrt{5}\)
b) \(\dfrac{x^2-y}{x-\sqrt{y}}=\dfrac{\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)}{x-\sqrt{y}}=x+\sqrt{y}\)
LC
2
24 tháng 8 2023
a: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)
b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)
c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
a) \(\dfrac{6}{5\sqrt{8}}\)
\(=\dfrac{6}{5\cdot2\sqrt{2}}\)
\(=\dfrac{6}{10\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{10}\)
b) \(\dfrac{7}{5+2\sqrt{3}}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}\)
\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)
\(=\dfrac{35-14\sqrt{3}}{13}\)
c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)
\(=3\sqrt{7}+3\sqrt{5}\)
7 tháng 7 2021
a) \(\dfrac{7}{\sqrt{5}-\sqrt{3}-\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}+\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{8-2\sqrt{15}-7}\)
\(=\dfrac{7\sqrt{5}-7\sqrt{3}+7\sqrt{7}}{1-2\sqrt{15}}\)
\(=\dfrac{\left(7\sqrt{5}-7\sqrt{3}+7\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{7\sqrt{5}+70\sqrt{3}-7\sqrt{3}-42\sqrt{5}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{-35\sqrt{5}+63\sqrt{3}+7\sqrt{7}+14\sqrt{105}}{-59}\)
\(=\dfrac{35\sqrt{5}-63\sqrt{3}-7\sqrt{7}-14\sqrt{105}}{59}\)
\(=\frac{6\sqrt{2}\left(3\sqrt{7}-5\sqrt{2}\right)}{2\left(3\sqrt{7}-5\sqrt{2}\right)}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\)