Tìm giá trị nhỏ nhất của biểu thức A= x + 1 + \(\dfrac{1}{x-1}\) biết rằng x > 1
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DT
5 tháng 4 2018
a) P=A:B \(\Leftrightarrow\dfrac{\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}+1}{x-1}\right)}{\dfrac{x}{\sqrt{x}-1}}\)
\(\Leftrightarrow\left(\dfrac{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}-1}{x}\)
\(\Leftrightarrow\left(\dfrac{x\sqrt{x}+x-\sqrt{x}-1-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{x}\right)\)
\(\Leftrightarrow\left(\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{x}\right)\)
\(\Leftrightarrow\dfrac{x-\sqrt{x}-2}{x\cdot\left(\sqrt{x}+1\right)}\)
b) để P < -1 thì :
\(\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+1\right)}< -1\)
\(\Rightarrow x-\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow x-\sqrt{x}+\sqrt{x}< -1+2\)
\(\Leftrightarrow x< 1\)
vậy,.............................
27 tháng 11 2018
ĐK:x>0
a) \(I=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\left(2\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}+1\right)+1-2\sqrt{x}-1=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)b)
Ta có \(I=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow x+\sqrt{x}-2\sqrt{x}-2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=0\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4 thì I=2
c)
Ta có x>1\(\Leftrightarrow x>\sqrt{x}\Leftrightarrow x-\sqrt{x}>0\)
Vậy \(I-\left|I\right|=x-\sqrt{x}-\left|x-\sqrt{x}\right|=x-\sqrt{x}-\left(x-\sqrt{x}\right)=0\)
d)\(I=x-\sqrt{x}=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{1}{4}\Leftrightarrow I\ge\dfrac{1}{4}\)
Dấu bằng xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy GTNN của I là \(\dfrac{1}{4}\) và xảy ra khi \(x=\dfrac{1}{4}\)
26 tháng 5 2022
a: \(A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)
Dấu '=' xảy ra khi x=-1/6
b: \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)
Dấu '=' xảy ra khi 4/9x-2/15=0
hay x=2/15:4/9=2/15x9/4=18/60=3/10
25 tháng 11 2022
a: \(P=\left(\dfrac{\sqrt{x}}{x+1}+\dfrac{1}{x+1}\right)\cdot\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Khi \(x=\dfrac{2+\sqrt{3}}{2}=\dfrac{4+2\sqrt{3}}{4}\) thì
\(P=\left(\dfrac{\sqrt{3}+1}{2}+1\right):\left(\dfrac{\sqrt{3}+1}{2}-1\right)\)
\(=\dfrac{\sqrt{3}+3}{2}:\dfrac{\sqrt{3}-1}{2}=\dfrac{3+\sqrt{3}}{\sqrt{3}-1}\)
15 tháng 4 2022
\(\left(x+1\right)\left(x+2\right)>=\left(x-2\right)^2-1\)
\(\Leftrightarrow x^2+3x+2>=x^2-4x+4-1\)
=>3x+2>=-4x+3
=>7x>=1
hay x>=1/7
21 tháng 6 2021
\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 1 2019
\(T=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{1}{x\sqrt{y}+y\sqrt{x}}\)
\(\Rightarrow T\ge\dfrac{1}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\ge\dfrac{1}{\dfrac{\left(x+y\right)}{2}.\sqrt{2\left(x+y\right)}}=\sqrt{2}\)
\(\Rightarrow T_{min}=\sqrt{2}\) khi \(x=y=\dfrac{1}{2}\)
TN
28 tháng 7 2017
a,\(P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
\(b,\) Để \(P=-\dfrac{1}{2}\) hay \(\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
\(\Leftrightarrow2x^2=-\left(x-1\right)\)
\(\Leftrightarrow2x^2=-x+1\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow2x^2+2x-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(A=x+1+\dfrac{1}{x-1}\\ \\ =x-1+2+\dfrac{1}{x-1}\\ =\left(x-1\right)+\dfrac{1}{x-1}+2\)
Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow A=\left(x-1\right)+\dfrac{1}{x-1}+2\ge2+2\ge4\)
Dấu "=" xảy ra khi:
\(x-1=1\\ \Leftrightarrow x=2\)
Vậy \(A_{Min}=4\) khi \(x=2\)