Rút gọn biểu thức A=\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) với a>0
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13 tháng 7 2019
a) ĐKXĐ : \(a>0;a\ne1\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right)\)
\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)
\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)
b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
LT
0
26 tháng 6 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x\left(\sqrt{x}-1\right)}\right):\left(\frac{1-\sqrt{x}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}-\frac{2}{1-x}\right)\)
\(=\left(\frac{x.\sqrt{x}}{x.\left(\sqrt{x}-1\right)}-\frac{1}{x\left(\sqrt{x}-1\right)}\right):\left(\frac{1-\sqrt{x}}{1-x}-\frac{2}{1-x}\right)\)
\(=\frac{x.\sqrt{x}-1}{x\left(\sqrt{x}-1\right)}.\frac{1-x}{-\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x.\sqrt{x}-1\right)\left(1-x\right)}{x\left(1-x\right)}=\frac{\sqrt{x^3}-1}{x}\)
\(b,\)\(A=\frac{\sqrt{x}^3-1}{x}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}\)
Để A > 0 \(\Rightarrow\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x}>0\)
Mà \(x>0\)và \(x+\sqrt{x}+1>0\)( do x lớn hơn 0 )
\(\Rightarrow\sqrt{x}-1>0\)
\(\Rightarrow\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1\)
A
20 tháng 6 2021
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\times\frac{\sqrt{x^2-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
Thay \(x=\frac{a^2+b^2}{2ab}\)vào A, ta được :
\(A=\frac{\sqrt{\frac{a^2+b^2}{2ab}+1}+\sqrt{\frac{a^2+b^2}{2ab}-1}}{\sqrt{\frac{a^2+b^2}{2ab}+1}-\sqrt{\frac{a^2+b^2}{2ab}-1}}\)
\(A=\frac{\sqrt{\frac{\left(a+b\right)^2}{2ab}}+\sqrt{\frac{\left(b-a\right)^2}{2ab}}}{\sqrt{\frac{\left(a+b\right)^2}{2ab}}-\sqrt{\frac{\left(b-a\right)^2}{2ab}}}\)
\(A=\frac{a+b\sqrt{\frac{1}{2ab}}+\left(b-a\right)\sqrt{\frac{1}{2ab}}}{a+b\sqrt{\frac{1}{2ab}}-\left(b-a\right)\sqrt{\frac{1}{2ab}}}\)
\(A=\frac{a+b+b-a}{a+b-b+a}\)
\(A=\frac{2b}{2a}\)
\(A=\frac{b}{a}\)
Ps : Nhớ k cho tui nhó, tui đã rất cố gắng rồi đấy. :)) K để lần sau có j tui giải giúp cho :)))
# Aeri #
16 tháng 10 2016
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
PN
2
J
8 tháng 6 2015
\(\sqrt{1+\left(\frac{1}{a}-\frac{1}{a+1}\right)^2+\frac{2}{a\left(a+1\right)}}=\sqrt{\left(\frac{1}{a\left(a+1\right)}\right)^2+\frac{2}{a\left(a+1\right)}+1}=\sqrt{\left(\frac{1}{a\left(a+1\right)}+1\right)^2}=\frac{1}{a\left(a+1\right)}+1=\frac{a^2+a+1}{a^2+a}\left(do\right)a>0\)
7 tháng 6 2015
=\(1+\frac{1}{a}+\frac{1}{a+1}\)
=\(\frac{a+1}{1}-\frac{1}{a+1}\)
18 tháng 5 2021
a,Với \(a>0;a\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{\sqrt{a}-1+a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{a-1}{a+\sqrt{a}}\)
b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)
\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)
Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)
13 tháng 8 2020
a) P = \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2.\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
P = \(\left(\frac{\sqrt{a}.\sqrt{a}-1}{2\sqrt{a}}\right)^2\cdot\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
P = \(\frac{\left(a-1\right)^2}{4a}\cdot\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
P = \(\frac{a-1}{4\sqrt{a}^2}\cdot\left(-4\sqrt{a}\right)\)
P = \(\frac{1-a}{\sqrt{a}}\)
b) với x > 0 và x khác 1
P < 0 => \(\frac{1-a}{\sqrt{a}}< 0\)
Do \(\sqrt{a}>0\) => 1 - a < 0 => a > 1
Vậy S = {a|a > 1}
HH
13 tháng 8 2020
Có 1 kiểu hơi khác Conan 1 tí -.-
\(a)P=\left(\frac{\sqrt{a}.\sqrt{a}-1}{2\sqrt{a}}\right).\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2.\frac{a-2\sqrt{a}+1-a-2\sqrt{1}-1}{a-1}=\frac{\left(a-1\right)\left(-4\sqrt{a}\right)}{\left(2\sqrt{a}\right)^2}\)
\(=\frac{\left(1-a\right).4\sqrt{a}}{4a}=\frac{1-a}{\sqrt{a}}\)
Vậy \(P=\frac{1-a}{\sqrt{a}}\)với a > 0 và \(a\ne1\)
b) Do a > 0 và a khác 1 nên P < 0 khi và chỉ khi :
\(\frac{1-a}{\sqrt{a}}< 0\Leftrightarrow1-a< 0\Leftrightarrow a>1\)
Có: \(A=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}\)
Có: \(1+a+\left(-a-1\right)=1+a-1-a=0\)
=> \(\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}=\sqrt{\left(\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\right)^2}=\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\)
=> \(A=1+\frac{1}{a}-\frac{1}{a+1}=1+\frac{1}{a\left(a+1\right)}\)
VẬY \(A=1+\frac{1}{a\left(a+1\right)}\)
\(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{a+1}\right)^2+\frac{2}{a\left(a+1\right)}+1}\)
\(=\sqrt{\left[\frac{1}{a\left(a+1\right)}+1\right]^2}=\left|\frac{1}{a}-\frac{1}{a+1}+1\right|\)