Cho A=\(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\). Khi \(x=\frac{\sqrt{3}}{4}\)là
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
30 tháng 7 2016
Xét : \(1+2x=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(1-2x=1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{4}\)
Ta có : \(A=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\sqrt{\left(\frac{\sqrt{3}+1}{2}\right)^2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}=\frac{\left(\sqrt{3}+1\right)^2}{2\left(3+\sqrt{3}\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2\left(3-\sqrt{3}\right)}\)
\(=\frac{1}{2\sqrt{3}}\left(\frac{4+2\sqrt{3}}{\sqrt{3}+1}+\frac{4-2\sqrt{3}}{\sqrt{3}-1}\right)=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}-4+6-2\sqrt{3}+4\sqrt{3}+4-6-2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}}{2}=1\)
AH
Akai Haruma
Giáo viên
30 tháng 6
Lời giải:
Tại $x=\frac{\sqrt{3}}{4}$:
\(\sqrt{1+2x}=\sqrt{1+\frac{\sqrt{3}}{2}}=\sqrt{\frac{2+\sqrt{3}}{2}}=\sqrt{\frac{4+2\sqrt{3}}{4}}=\sqrt{\frac{(\sqrt{3}+1)^2}{2^2}}=\frac{\sqrt{3}+1}{2}\)
\(\sqrt{1-2x}=\sqrt{1-\frac{\sqrt{3}}{2}}=\sqrt{\frac{2-\sqrt{3}}{2}}=\sqrt{\frac{4-2\sqrt{3}}{4}}=\sqrt{\frac{(\sqrt{3}-1)^2}{2^2}}=\frac{\sqrt{3}-1}{2}\)
\(A=\frac{1+\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}+1}{2}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}-1}{2}}\\ =\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\\ =\frac{4+2\sqrt{3}}{2\sqrt{3}(\sqrt{3}+1)}+\frac{4-2\sqrt{3}}{2\sqrt{3}(\sqrt{3}-1)}\\ =\frac{(\sqrt{3}+1)^2}{2\sqrt{3}(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)^2}{2\sqrt{3}(\sqrt{3}-1)}\\ =\frac{\sqrt{3}+1}{2\sqrt{3}}+\frac{\sqrt{3}-1}{2\sqrt{3}}=\frac{2\sqrt{3}}{2\sqrt{3}}=1\)
DH
4 tháng 4 2020
\(a,M=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\frac{2x-2\sqrt{2}x+2\sqrt{2x}-1}{2x-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x+1}}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\frac{-2\sqrt{2}x+2\sqrt{2x}}{2x-1}\right):\left(1+\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\right)\)
\(=\left(\frac{-2\sqrt{2}x+2\sqrt{2x}}{2x-1}\right):\left(\frac{-2\sqrt{x}-2}{2x-1}\right)\)
\(=\frac{-\sqrt{2}x+\sqrt{2x}}{\sqrt{x}-1}\)
\(=\frac{-\sqrt{2x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=-\sqrt{2x}\)
\(b,x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)
\(x=\frac{1}{2}\left(1+2\sqrt{2}+2\right)\)
\(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\)
Thay \(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\) vào \(M=-\sqrt{2x}\) ta được:
\(M=-\sqrt{2.\frac{1}{2}\left(1+\sqrt{2}\right)^2}\)
\(M=-1-\sqrt{2}\)
Vậy ..............