MT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
7 tháng 11 2018
\(A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}+1}}=\dfrac{1-\sqrt{x-1}}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)
\(A=\dfrac{1-\sqrt{x-1}}{\left|\sqrt{x-1}-1\right|}\) \(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}=-1\\A=\dfrac{1-\sqrt{x-1}}{-\sqrt{x-1}+1}=1\end{matrix}\right.\)
7 tháng 11 2018
\(B=\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(B^2=\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)^2\)
\(B^2=8+2\sqrt{10+2\sqrt{5}}-2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}+8-2\sqrt{10+2\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(B^2=16-2\sqrt{8^2-4\left(10+2\sqrt{5}\right)}\)
\(B^2=16-2\sqrt{24-8\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(B^2=16-4\sqrt{5}+4=20-4\sqrt{5}\)
\(B=\sqrt{20-4\sqrt{5}}\)
17 tháng 8 2017
c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)
=> đpcm.
26 tháng 8 2023
2:
ĐKXĐ: x>=3
\(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)
=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)
TH1: x>=6
(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)
=>\(2\sqrt{x-3}=2\sqrt{3}\)
=>x-3=3
=>x=6(nhận)
TH2: 3<=x<6
Phương trình (1) sẽ là;
\(\sqrt{x-3}+\sqrt{3}-\sqrt{x-3}=\sqrt{3}\)
=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)
1:
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)
\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)
\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)
\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)
\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)
=>\(A=\sqrt{10}+\sqrt{2}\)
8 tháng 7 2021
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
8 tháng 7 2021
Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
DH
1
VA
1
24 tháng 2 2020
\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\) \(=\frac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)
=\(2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
8 tháng 8 2022
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
mình chép ra giấy
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}+8-2\sqrt{10+2\sqrt{5}}\)
\(=16+2\sqrt{24-2.2.2\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\left(2\sqrt{5}-2\right)=16+4\sqrt{5}-4=12+4\sqrt{5}\)
Vậy \(A=\sqrt{12+4\sqrt{5}}=\sqrt{4.3+4\sqrt{5}}=2\sqrt{3+\sqrt{5}}\)