Chứng minh: \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\), với mọi a > 1
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
NM
1
22 tháng 10 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(A=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{x-1}{x-\sqrt{x}}\cdot\left(\sqrt{x}+1\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi \(x=\left(\sqrt{3}-1\right)^2\) thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\left(\sqrt{3}+1\right)}{2}=\dfrac{3\sqrt{3}+3}{2}\)
c: \(P-2=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}-2\)
\(=\dfrac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}>0\)
=>P>2
HS
21 tháng 8 2018
\(\dfrac{1+a}{1+\sqrt{a}+1}=\dfrac{\sqrt{6}}{1+\sqrt{6}}\\ \Leftrightarrow\left(1+\sqrt{6}\right)\left(1+a\right)=\sqrt{6}\left(1+\sqrt{a}+a\right)\\ \Leftrightarrow1+a+\sqrt{6}+\sqrt{6}a=\sqrt{6}+\sqrt{6}a+\sqrt{6}a\\ \Leftrightarrow1+a=\sqrt{6}a\\ \Leftrightarrow\sqrt{6}a=1+a\\ \Leftrightarrow6a=1+2a+a^2\\ \Leftrightarrow6a-1-2a-a^2=0\\ \Leftrightarrow4a-1-a^2=0\\ \Leftrightarrow-a^2+4a-1=0\\ \Leftrightarrow a^2-4a+1=0\)
\(a=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4.1.1}}{2.1}\\ a=\dfrac{4\pm\sqrt{16-4}}{2}\\ a=\dfrac{4\pm\sqrt{12}}{2}\\ a=\dfrac{4\pm2\sqrt{3}}{2}\)
\(\left[{}\begin{matrix}a=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\a=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\end{matrix}\right.\)
Vậy......
Sr bạn cách làm mình hơi khó hiểu chút :v
HT
21 tháng 8 2018
a)ĐK: \(a\ge0\)
\(\Leftrightarrow\left(1+a\right)\left(1+\sqrt{6}\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\)
\(\Leftrightarrow1+\sqrt{6}+a+a\sqrt{6}=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)
\(\Leftrightarrow1+a=\sqrt{6a}\)
\(\Leftrightarrow a^2+2a+1-6a=0\)
\(\Leftrightarrow a^2-4a+1=0\)
\(\Leftrightarrow\left(a-2\right)^2-3=0\)
\(\Leftrightarrow\left(a-2\right)^2=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-2=\sqrt{3}\\a-2=-\sqrt{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\sqrt{3}+2\\a=2-\sqrt{3}\end{matrix}\right.\)
b)
MK sẽ chứng minh tương đương:
\(\Leftrightarrow\dfrac{1+a}{1+\sqrt{a}+a}-\dfrac{2}{3}>0\)
\(\Leftrightarrow\dfrac{a-2\sqrt{a}+1}{1+\sqrt{a}+a}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)
Ta có:
\(\left(\sqrt{a}-1\right)^2\ge0\left(1\right)\)
\(1+\sqrt{a}+a=\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà \(\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Leftrightarrow1+\sqrt{a}+a>0\left(2\right)\)
Từ (1), (2)
=>\(\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)
=>\(\dfrac{1+a}{1+\sqrt{a}+a}>\dfrac{2}{3}\)
20 tháng 12 2021
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)
14 tháng 8 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
14 tháng 8 2021
c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)
14 tháng 4 2023
a: \(A=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: \(A-5=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>=0\)
=>A>=5
\(\sqrt{a+1}-\sqrt{a-1}=\dfrac{\left(\sqrt{a+1}-\sqrt{a-1}\right)\left(\sqrt{a+1}+\sqrt{a-1}\right)}{\sqrt{a+1}+\sqrt{a-1}}\)
\(=\dfrac{2}{\sqrt{a+1}+\sqrt{a-1}}\)
Mà \(1.\sqrt{a+1}+1.\sqrt{a-1}< \sqrt{\left(1+1\right)\left(a+1+a-1\right)}=2\sqrt{a}\) (dấu "=" của BĐT Bunhia ko xảy ra)
\(\Rightarrow\dfrac{2}{\sqrt{a+1}+\sqrt{a-1}}>\dfrac{2}{2\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)
Hay \(\dfrac{1}{\sqrt{a}}< \sqrt{a+1}-\sqrt{a-1}\) (đpcm)