A = \(\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x^2}{x\sqrt{x}-x}\right)\left(2-\frac{1}{\sqrt{x}}\right)\))
a, RG
b, tìm x để A =0, A= 3
c, tìm x để A +|A| = 0
K
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NV
13 tháng 7 2016
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
4 tháng 8 2019
\(A=\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}-1}\right)\)\(:\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{\left(\sqrt{x}-1-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}-4-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(b,A=0\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=0\Leftrightarrow\sqrt{x}-2=0\)
Mà \(\sqrt{x}+2\ne0\)\(\Rightarrow\)không có giá trị nào của x thỏa mãn \(A=0\)
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 )
\(\Rightarrow\sqrt{x}-1>0\)
\(\Rightarrow\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1\)
\(A=\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x^2}{x\sqrt{x}-x}\right)\left(2-\frac{1}{\sqrt{x}}\right)\left(ĐKXĐ:0< x;x\ne1\right)\)
\(A=\left(\frac{x^2\sqrt{x}}{x\left(\sqrt{x}-1\right)}-\frac{x^2}{x\left(\sqrt{x}-1\right)}\right)\left(\frac{2\sqrt{x}-1}{2\sqrt{x}}\right)\)
\(A=\left(\frac{x^2\left(\sqrt{x}-1\right)}{x\left(\sqrt{x}-1\right)}\right)\left(\frac{2\sqrt{x}-1}{2\sqrt{x}}\right)\)
\(A=x.\left(\frac{2\sqrt{x}-1}{2\sqrt{x}}\right)\)
\(A=\frac{x\left(2\sqrt{x}-1\right)}{2\sqrt{x}}\)
b)Tại A=0(ĐKXĐ:0<x;x khác 1) ta đc:
\(A=\frac{x\left(2\sqrt{x}-1\right)}{2\sqrt{x}}=0\)
\(\Leftrightarrow x\left(2\sqrt{x}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{x}-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\left(kOTM\right)\\x=\frac{1}{4}\end{cases}}\)
Vậy tại A=0 x=1/4
Tại A=3(ĐKXĐ:0<x;x khác 1) ta đc:
\(\frac{x\left(2\sqrt{x}-1\right)}{2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}^3-x=6\sqrt{x}\)
\(\Leftrightarrow x=0\left(koTM\right)\)