Rút gọn biểu thức A = \(\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}\)
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
23 tháng 4 2021
\(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{3-\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{3+\sqrt{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)
23 tháng 4 2021
a, \(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt[]{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)
7 tháng 10 2016
\(A=\frac{\sqrt{3}-1}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\sqrt{3}-1}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\sqrt{3}-1}{1+\frac{\sqrt{4+2\sqrt{3}}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\sqrt{3}-1}{\frac{3+\sqrt{3}}{2}}+\frac{\sqrt{3}+1}{\frac{3-\sqrt{3}}{2}}=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\frac{2}{\sqrt{3}}\left(\frac{4-2\sqrt{3}+4+2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)=\frac{2}{\sqrt{3}}.\frac{8}{2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
MU
3
5 tháng 7 2019
\(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}\)
\(=\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{3+\sqrt{7}-3+\sqrt{7}}{9-7}\)
\(=\frac{2\sqrt{7}}{2}\)
\(=\sqrt{7}\)
phần B là gì cơ?
\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{3^2-\sqrt{2}^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\sqrt{3}^2}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)
\(=3-\sqrt{2}-1-\sqrt{3}\)
\(=2-\sqrt{2}-\sqrt{3}\)
\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}=\frac{7\left(3-\sqrt{2}\right)}{3^2-\left(\sqrt{2}\right)^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\left(\sqrt{3}\right)^2}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{9-2}+\frac{2\left(1+\sqrt{3}\right)}{1-3}=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)
\(=\left(3-\sqrt{2}\right)-\left(1+\sqrt{3}\right)=3-\sqrt{2}-1-\sqrt{3}=2-\sqrt{2}-\sqrt{3}\)