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12 tháng 11 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để A<0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<x<1
H
12 tháng 11 2023
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\left(đkxđ:x>0;x\ne1\right)\\ =\left(\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}\left(\sqrt{x-1}\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\\ =\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
`b,` Để `A<0` thì :
\(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.2\left(\sqrt{x}+1\right)>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)
Kết hợp với điều kiện xác định ta có : \(0< x< 1\)
12 tháng 4 2021
ĐKXĐ : x > 0 , x khác 1
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\div\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\div\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\div\left[\dfrac{x+\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=2\div\dfrac{1}{\sqrt{x}-1}=2\sqrt{x}-2\)
b) Dễ thấy ∀ x ≥ 0 thì \(2\sqrt{x}-2\) nguyên
Kết hợp với ĐKXĐ => Với \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)thì A đạt giá trị nguyên
AA
6 tháng 8 2021
a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)
A=\(\frac{1}{x+2\sqrt{x}}\)
b, A >= \(\frac{1}{3\sqrt{x}}\)
=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)
=> x <= -1 , x >= 4 (x khác 0)
BT
27 tháng 9 2017
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
10 tháng 10 2021
\(P=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Rightarrow P=\dfrac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x-\sqrt{x}+x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}-x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x-\sqrt{x}+x\sqrt{x}-1-\left(x+\sqrt{x}-x\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x^2-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x-\sqrt{x}+x\sqrt{x}-1-x-\sqrt{x}+x\sqrt{x}+1+x^2-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{x^2-2\sqrt{x}+2x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\dfrac{3-x+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(=\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}=\dfrac{3}{\sqrt{x}-2}\)
b: A<1
=>A-1<0
=>\(\dfrac{3-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{\sqrt{x}-5}{\sqrt{x}-2}>0\)
TH1: \(\left\{{}\begin{matrix}\sqrt{x}-5>0\\\sqrt{x}-2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}>5\\\sqrt{x}>2\end{matrix}\right.\Leftrightarrow\sqrt{x}>5\)
=>x>25
TH2: \(\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}-2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 5\end{matrix}\right.\)
=>\(\sqrt{x}< 2\)
=>0<=x<4