Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2.
a, \(P=\left(\frac{a\sqrt{a}+1}{a-1}-\frac{a-1}{\sqrt{a}-1}\right):\left(\sqrt{a}-\frac{\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left[\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{a-1}{\sqrt{a}-1}\right]:\frac{a-\sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\)
\(=\left[\frac{a-\sqrt{a}+1}{\sqrt{a}-1}-\frac{a-1}{\sqrt{a}-1}\right]:\frac{a-2\sqrt{a}}{\sqrt{a}-1}\)
\(=\frac{2-\sqrt{a}}{\sqrt{a}-1}.\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-2\right)}=-\frac{1}{\sqrt{a}}\)
b, \(a=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{a}=\sqrt{2}-1\)
Khi đó \(P=-\frac{1}{\sqrt{a}}=-\frac{1}{\sqrt{2}-1}=-\sqrt{2}-1\)
1.
a, \(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(=a-\sqrt{a}\)
b, \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(\Rightarrow MinA=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
\(C=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{a-\sqrt{a}}\right):\frac{\sqrt{a}-1}{a}\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}-1}{a}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}-1}{a}=\frac{a-1}{\sqrt{a}}:\frac{\sqrt{a}-1}{a}=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{a}{\sqrt{a}-1}\)
\(=\left(\sqrt{a}+1\right)\sqrt{a}\)
Với a > 0 , a khác 1 ta có :
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\cdot\frac{a}{\sqrt{a}-1}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\frac{a}{\sqrt{a}-1}=\frac{a-1}{\sqrt{a}}\cdot\frac{a}{\sqrt{a}-1}=\sqrt{a}\left(\sqrt{a}+1\right)=a+\sqrt{a}\)
*bài này làm sao ấy tìm Min không ra:v*
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) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
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 \(M< 1\)
a) \(P=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\frac{\sqrt{a}.\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}.\left(\sqrt{a}+1\right).\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1=a-\sqrt{a}\)
b)Ta có a>0 do đó: \(P=a-\sqrt{a}\ge0\)
Dấu "=" xảy ra khi a=1
c) Ta thấy \(P\ge0\)
=>P2\(\ge\)P
=>P\(\ge\)\(\sqrt{P}\)
ĐKXĐ: \(a>0\)
a/ \(P=\frac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
b/ Ta có: \(\hept{\begin{cases}a>0\\\sqrt{a}\ge0\end{cases}\Rightarrow a-\sqrt{a}\ge0}\)
MinP = 0 khi \(\sqrt{a}=0\Rightarrow a=0\)
c/ \(P\ge\sqrt{P}\)