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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Xét f(u) = \(\dfrac{u^p}{p}+\dfrac{v^q}{q}-uv,u\ge0\)
( Xem v > 0 vì v = 0 : BĐT luôn đúng )
f '(u) = up-1 - v = 0 \(\Leftrightarrow\) up-1 = v \(\Leftrightarrow\) u = \(v^{\dfrac{q}{p}}\)
Vẽ bảng biến thiên ( tự vẽ )
Vậy \(uv\le\dfrac{u^p}{p}+\dfrac{v^q}{q}\)
b)* Nếu \(\int\limits^b_a\left|f\left(x\right)\right|^pdx=0\) hay \(\int\limits^b_a\left|g\left(x\right)\right|^qdx=0\)thì \(f\equiv0\)hay \(g\equiv0\) BĐT luôn đúng
Xét \(\int\limits^b_a\left|f\left(x\right)\right|^pdx>0\) và \(\int\limits^b_a\left|g\left(x\right)\right|^qdx>0\)
Áp dụng BĐT câu (a) :
Với \(\left\{{}\begin{matrix}u=\dfrac{\left|f\left(x\right)\right|}{\left(\int\limits^b_a\left|f\left(x\right)\right|^pdx\right)^{\dfrac{1}{p}}}>0\\v=\dfrac{\left|g\left(x\right)\right|}{\left(\int\limits^b_a\left|g\left(x\right)\right|^qdx\right)^{\dfrac{1}{q}}}>0\end{matrix}\right.\)
\(uv\le\dfrac{u^p}{p}+\dfrac{v^q}{q}\left(1\right)\)
Lấy tích phân từ a \(\rightarrow\) b 2 vế BĐT (1) ta được :
\(\int\limits^b_auvdx\le\dfrac{1}{p}+\dfrac{1}{q}=1\)
Vậy : \(\int\limits^b_a\left|f\left(x\right).g\left(x\right)\right|dx\le\left(\int\limits^b_a\left|f\left(x\right)^p\right|dx\right)^{\dfrac{1}{p}}\left(\int\limits^b_a\left|g\left(x\right)^q\right|dx\right)^{\dfrac{1}{q}}\)
\(\Rightarrow\)(Đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
1)\(\forall x1,x2\in\left(1,+\infty\right),x1\ne x2\)
\(f\left(x1\right)-f\left(x2\right)=\dfrac{1}{1-x1}-\dfrac{1}{1-x2}=\dfrac{1-x2-1+x1}{\left(1-x1\right)\left(1-x2\right)}=\dfrac{x1-x2}{\left(1-x1\right)\left(1-x2\right)}\)
\(\dfrac{f\left(x1\right)-f\left(x2\right)}{x1-x2}=\dfrac{\dfrac{x1-x2}{\left(1-x1\right)\left(1-x2\right)}}{x1-x2}=\dfrac{1}{\left(1-x1\right)\left(1-x2\right)}\)
vì \(x1,x2\in\left(1;+\infty\right)\)nên \(\left\{{}\begin{matrix}x1>1\\x2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-x1< 0\\1-x2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{\left(1-x1\right)\left(1-x2\right)}>0\)
Vậy hàm số đồng biến trên \(\left(1;+\infty\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lấy x1;x2<1 sao cho x1<x2
\(A=\dfrac{f\left(x1\right)-f\left(x2\right)}{x_1-x_2}=\left(\dfrac{x_1-2}{x_1+1}-\dfrac{x_2-2}{x_2+1}\right):\left(x_1-x_2\right)\)
\(=\dfrac{x_1x_2+x_1-2x_2-2-x_1x_2-x_2+2x_1+2}{\left(x_1+1\right)\left(x_2+1\right)}\cdot\dfrac{1}{x_1-x_2}\)
\(=\dfrac{3x_1-3x_2}{\left(x_1+1\right)\left(x_2+1\right)}\cdot\dfrac{1}{x_1-x_2}=\dfrac{3}{\left(x_1+1\right)\left(x_2+1\right)}\)
x1<-1; x2<-1 nên x1+1<0; x2+1<0
=>(x1+1)(x2+1)>0
=>A>0
=>Hàm số đồng biến khi x<-1
Khi x1>-1; x2>-1 thì x1+1>0; x2+1>0
=>(x1+1)(x2+1)>0
=>A>0
=>Hàm số đồng biến khi x>-1
=>Hàm số đồng biến khi x<>-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(f\left(x\right)=\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}.\)
\(x-1=0.\Leftrightarrow x=1.\\ x-2=0.\Leftrightarrow x=2.\\ x-3=0.\Leftrightarrow x=3.\)
\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\) \(\left(1;2\right)\cup\left(3;+\infty\right).\)
\(\Rightarrow B.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{4}{x_1+1}-\dfrac{4}{x_2+1}}{x_1-x_2}\)
\(=\dfrac{4x_2+4-4x_1-4}{\left(x_1+1\right)\left(x_2+1\right)}\cdot\dfrac{1}{x_1-x_2}=\dfrac{-4}{\left(x_1+1\right)\left(x_2+1\right)}\)
Nếu \(x\in\left(-\infty;-1\right)\) thì \(\left\{{}\begin{matrix}x_1< -1\\x_2< -1\end{matrix}\right.\Leftrightarrow\left(x_1+1\right)\left(x_2+1\right)>0\)
=>A<0
=>f(x) nghịch biến trên khoảng \(\left(-\infty;-1\right)\)
Nếu \(x\in\left(-1;+\infty\right)\) thì \(\left\{{}\begin{matrix}x_1>-1\\x_2>-1\end{matrix}\right.\Leftrightarrow\left(x_1+1\right)\left(x_2+1\right)>0\)
=>A<0
=>f(x) nghịch biến trên khoảng \(\left(1;+\infty\right)\)