Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ĐKXĐ:x\ne1\)
a) \(Q=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}+\frac{3-\sqrt{x}}{x-1}\)
\(=\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)+\sqrt{x}-3}{1-x}\)
\(=\frac{\sqrt{x}+x+\sqrt{x}-x+\sqrt{x}-3}{1-x}\)
\(=\frac{3\sqrt{x}-3}{1-x}=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-3}{\sqrt{x}+1}\)
b) Ta có \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}+1\ge1\)
\(\Rightarrow\frac{-3}{\sqrt{x}+1}\ge-3\)
Dấu "=" khi x = 0
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
Bạn tự tìm ĐKXĐ nhé :)
Xét tử thức : \(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
Xét mẫu thức : \(\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}=\sqrt{\left(\frac{4}{x}-1\right)^2}=\left|\frac{4}{x}-1\right|=\left|\frac{x-4}{x}\right|\)
Từ đó rút gọn P
1) \(\sqrt{4+x}=2-x\) (ĐK: \(x\ge-4\))
\(\Leftrightarrow4+x=\left(2-x\right)^2\)
\(\Leftrightarrow4+x=4-4x+x^2\)
\(\Leftrightarrow x^2-4x-x+4-4=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{0;5\right\}\)
2)
a) ĐKXĐ: \(a>0,a\ne1\)
\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)
\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]\cdot\dfrac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)
\(A=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\cdot\dfrac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}+1}\)
\(A=\sqrt{a}\left(\sqrt{a}-1\right)\)
\(A=a-\sqrt{a}\)
b) Ta có:
\(A=a-\sqrt{a}\)
\(A=\left(\sqrt{a}\right)^2-2\cdot\dfrac{1}{2}\cdot\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\)
\(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Mà: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi:
\(\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=-\dfrac{1}{4}\)
\(\Leftrightarrow a=\dfrac{1}{4}\)
Vậy: \(A_{min}=-\dfrac{1}{4}\)khi \(a=\dfrac{1}{4}\)