Tìm GTNN và giá trị tương ứng của x trong: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
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Những câu hỏi liên quan
19 tháng 6 2023
a: ĐKXĐ: x>0; x<>1
b: \(A=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
c: A nguyên
=>x-1 thuộc {1;-1;2;-2}
=>x thuộc {2;3}
MN
30 tháng 3 2020
ĐK: \(x\ge0\)
\(M=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
Để M min \(\Leftrightarrow\frac{3}{\sqrt{x}+2}\)max
\(\Leftrightarrow\sqrt{x}+2\)min
\(\Leftrightarrow\sqrt{x}\)min
\(\Leftrightarrow x\)min
mà \(x\ge0\)
\(\Leftrightarrow x=0\)
Vậy \(Min_M=\frac{-1}{2}\Leftrightarrow x=0\)
NV
14 tháng 7 2016
a/ \(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
+ Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
+ Với \(\sqrt{x}+1=-1\Rightarrow\sqrt{x}=-2\left(vn\right)\)
+ Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)
+ Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)
Vậy x = 0 thì P nguyên
14 tháng 7 2016
a) \(P=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)
\(=\frac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)
c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0
Vậy Min P = 1/2 <=> x = 0
29 tháng 10 2023
a: \(B=\dfrac{\sqrt{x}}{x+\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x+1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: B=2/7
=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)
=>\(2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)
=>\(2x+2\sqrt{x}-7\sqrt{x}+2=0\)
=>\(2x-5\sqrt{x}+2=0\)
=>\(\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
30 tháng 6 2021
*Rút gọn
Ta có: \(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
Ta có: \(C=x-\sqrt{x}+1\)
\(=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\)
hay \(x=\dfrac{1}{4}\)
30 tháng 6 2021
\(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(C_{min}=\dfrac{3}{4}\)
\(N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}\)
Áp dụng AM-GM có: \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\)
Dấu "=" xảy ra khi x=1 (ktm đk)
Suy ra dấu bằng ko xảy ra \(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}< 2\)
\(\Rightarrow N< 2\) mà \(N>0\),\(N\) nguyên
\(\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\) (tm)
Vậy...
TM
26 tháng 6 2023
Ta có : \(P=3A+2B\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{3}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}.\)
\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+2\right)-1}{\sqrt{x}+2}=2-\dfrac{1}{\sqrt{x}+2}\)
Do \(x\ge0\Rightarrow\sqrt{x}+2\ge0\)
\(\Rightarrow-\dfrac{1}{\sqrt{x}+2}\ge-1\)
\(\Rightarrow P=2-\dfrac{1}{\sqrt{x}+2}\ge-1+2=1.\)
Vậy : \(MinP=1.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=0.\)
2
14 tháng 5 2023
`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`
`A=4/[12-4]=1/2`
Với `x > 0,x ne 1` có:
`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`
`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`
`b)B=2/5A`
`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`
`<=>5\sqrt{x}-5=\sqrt{x}`
`<=>\sqrt{x}=5/4`
`<=>x=25/16` (t/m)
4 tháng 8 2017
a) \(U=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(U=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(U=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(U=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(U=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(U=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\Leftrightarrow\dfrac{-5\left(\sqrt{x}-1\right)\left(\sqrt{x}-\dfrac{2}{5}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{-5\left(\sqrt{x}+\dfrac{2}{5}\right)}{\sqrt{x}+3}\Leftrightarrow\dfrac{-5\sqrt{x}-2}{\sqrt{x}+3}\)
b) ta có \(U=\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{-5\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{2}\) \(\Leftrightarrow\sqrt{x}+3=2\left(-5\sqrt{x}-2\right)\)
\(\Leftrightarrow\sqrt{x}+3=-10\sqrt{x}-4\Leftrightarrow\sqrt{x}+3-\left(-10\sqrt{x}-4\right)\)
\(\Leftrightarrow\sqrt{x}+3+10\sqrt{x}+4=0\Leftrightarrow11\sqrt{x}+7=0\Leftrightarrow11\sqrt{x}=-7\)
\(\Leftrightarrow\sqrt{x}=\dfrac{-7}{11}\left(vôlí\right)\)
vậy không có giá trị nào để \(U=\dfrac{1}{2}\)
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
Có \(\sqrt{x}+2\ge2\forall x\ge0\)
\(\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Rightarrow-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{3}{2}\)\(\Leftrightarrow1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\ge-\dfrac{1}{2}\)
Dấu "=" xảy ra khi x=0
Vậy \(min=-\dfrac{1}{2}\Leftrightarrow x=0\)