Bài 1 :
T =\(\left(\frac{\sqrt{x}+2}{3\sqrt{x}}+\frac{2}{\sqrt{x}+1}-3\right):\frac{2-4\sqrt{x}}{\sqrt{x}+1}+\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
a, Tìm ĐKXĐ
b, Rút gọn
c, Tìm x để T < 0
Help me !!!
K
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5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
31 tháng 7 2016
N=\(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right).\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)
ĐKXĐ \(\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\3-x\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\x\ne3\\x\ge0\end{cases}}\)
\(=\left[\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right].\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right).\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\frac{x-2x+3}{3-x}=\frac{3-x}{3-x}=1\)
31 tháng 7 2016
câu 2 ra |a-b| nha bn mik đăng rồi nhưng bị lỗi nên nó ko hiện lên
24 tháng 7 2017
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
a, ĐKXĐ: x>0 (1)
b,T= (\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)+6\sqrt{x}-9\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)).(\(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\))+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{x+3\sqrt{x}+2+6\sqrt{x}-9x-9\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2-8x}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2-4\sqrt{x}}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\left(\frac{2\left(1-2\sqrt{x}\right)\left(1+2\sqrt{x}\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\right)\).\(\left(\frac{\sqrt{x}+1}{2\left(1-2\sqrt{x}\right)}\right)\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}}{3\sqrt{x}}\)+\(\frac{x-3\sqrt{x}-1}{3\sqrt{x}}\) = \(\frac{x-\sqrt{x}}{3\sqrt{x}}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{3\sqrt{x}}\)=\(\frac{\sqrt{x}-1}{3}\)
c, Để T<0 \(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{3}\) <0 \(\Leftrightarrow\) \(\sqrt{x}\)-1<0 \(\Leftrightarrow\) \(\sqrt{x}\)<1\(\Leftrightarrow\) x<1 mà do ĐK (1)
=> Để T<0 \(\Leftrightarrow\) 0<x<1
Cho mk hỏi là bước t2 từ dưới lên phần b thì \(\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)\) sao lại khai triển đc như vậy