Cho A= \(\dfrac{x}{x-2}\) .
Tìm GTNN của bt x.A biết x>2.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Bunhiacopxki
\(\left(x+y\right)^2\le\left(x^2+y^2\right)\left(1+1\right)=2.2=4\)
<=>\(-2\le x+y\le2\)
GTNN của x+y là -2 khi x=y=-1
GTLN của x+y là 2 khi x=y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vì \(\left(2x+\frac{1}{4}\right)^4\ge0\forall x\)
\(\Rightarrow A\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow2x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{8}\)
b) \(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)
\(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)
Vì \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\forall x\)
\(\Rightarrow B\le3\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\)
với mọi x thì (2x+1/4)4>=0 (lớn hơn hoặc bằng )
A=(2x+1/4)4-1>=-1
để A đạt GTNN thì (2x+1/4)4=0
2x+1/4=0 =>x=-1/8
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x+\dfrac{1}{x-2}\\ \Rightarrow A=x-2+\dfrac{1}{x-2}+2\)
Áp dụng BĐT Cô-si ta có:
\(A=x-2+\dfrac{1}{x-2}+2\\ \ge2\sqrt{\left(x-2\right).\dfrac{1}{x-2}}+2\\ =2\sqrt{1}+2\\ =4\)
\(\text{Dấu "=" xảy ra}\Leftrightarrow x-2=\dfrac{1}{x-2}\\ \Leftrightarrow\left(x-2\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy \(A_{min}=4\Leftrightarrow x=3\)
\(A=x-2+\dfrac{1}{x-2}+2\ge2+2=4\)
Dấu '=' xảy ra khi x-2=1 hoặc x-2=-1
=>x=3 hoặc x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{x}{x-2}=>x.A=\dfrac{x.x}{x-2}=\dfrac{x.x-2.2+4}{x-2}\)
\(\Leftrightarrow x.A=x+2+\dfrac{4}{x-2}=\left(x-2\right)+\dfrac{4}{x-2}+4\)
có \(x>2\Leftrightarrow x-2>0\Rightarrow x-2=\sqrt{\left(x-2\right)^2}\)
\(x.A=\left(\sqrt{x-2}-\dfrac{2}{\sqrt{x-2}}\right)^2+8\)
có \(\left(\sqrt{x-2}-\dfrac{2}{\sqrt{x-2}}\right)^2\ge0\left\{x=4\right\}\)
GTNN x.A =8 khi x =4