Cho A=\(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\). Khi \(x=\frac{\sqrt{3}}{4}\)gía trị cuả A là?
CHỈ CHO MÌNH CÁCH LÀM VỚI
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bản rút gọn biểu thức trên A =\(x-\sqrt{x}+2\)
=\(x-2\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2\)
= \(\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}\)
vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)với mọi x
<=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)voi mọi x
<=> A \(\ge\)7/4
=> min A = 7/4
dau = xay ra <=> \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
\(ĐKXĐ:\) \(\hept{\begin{cases}\sqrt{x}-1\ne0\\\sqrt{x}\ge0\\x-\sqrt{x}+1\ne0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\) ( vì \(x-\sqrt{x}+1>0\) )
Ta có:
\(A=x-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1=x-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x^3}+1}{x-\sqrt{x}+1}+1\)
\(=x-2\sqrt{x}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1=x-2\sqrt{x}+\sqrt{x}+1+1\)
nên \(A=x-\sqrt{x}+2=x-2.\frac{1}{2}\sqrt{x}+\frac{1}{4}+\frac{7}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
Vậy, \(A_{min}=\frac{7}{4}\) khi \(x=\frac{1}{4}\)
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 : \(A=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\sqrt{\left(\frac{\sqrt{3}+1}{2}\right)^2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}=\frac{\left(\sqrt{3}+1\right)^2}{2\left(3+\sqrt{3}\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2\left(3-\sqrt{3}\right)}\)
\(=\frac{1}{2\sqrt{3}}\left(\frac{4+2\sqrt{3}}{\sqrt{3}+1}+\frac{4-2\sqrt{3}}{\sqrt{3}-1}\right)=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}-4+6-2\sqrt{3}+4\sqrt{3}+4-6-2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}}{2}=1\)
WhatTheFackNgaoVc