\(log_2\left(1+log_{3^{-2}}x-log_{3^2}x\right)< 1\)

\(\Leftrightarrow log_2\left(1-\dfrac{1}{2}log_3x-\dfrac{1}{2}log_3x\right)< 1\)

\(\Leftrightarrow log_2\left(1-log_3x\right)< 1\)

\(\Leftrightarrow0< 1-log_3x< 2\)

\(\Leftrightarrow-1< log_3x< 1\)

\(\Leftrightarrow\dfrac{1}{3}< x< 3\Rightarrow\left\{{}\begin{matrix}a=3\\b=3\end{matrix}\right.\) \(\Rightarrow a=b\)

Lời giải:

Ta có:

\(\text{VT}=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}\)

\(=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)

\(=(a+b+c)-2\left(\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\right)\)

Áp dụng BĐT Cauchy cho các số dương:

\(\text{VT}\geq (a+b+c)-2\left(\frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\right)\)

\(\Leftrightarrow \text{VT}\geq (a+b+c)-\frac{2}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)

Áp dụng BĐT Cauchy tiếp:

\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}\)

\(=\frac{2(ab+bc+ac)+3}{3}\leq \frac{2.\frac{(a+b+c)^2}{3}+3}{3}\)

Do đó: \(\text{VT}\geq (a+b+c)-\frac{2}{3}.\frac{2.\frac{(a+b+c)^2}{3}+3}{3}=1\) do $a+b+c=3$

Ta có đpcm

Dấu bằng xảy ra khi $a=b=c=1$

Phương trình đã cho tương đương:

\(\left\{{}\begin{matrix}x+2\ge0\\2x^2+mx+3=\left(x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+\left(m-4\right)x-1=0\end{matrix}\right.\)

Do \(a.c=1.\left(-1\right)=-1< 0\Rightarrow\) pt đã cho luôn có hai nghiệm \(x_1;x_2\). Ta tìm điều kiện m để \(-2\le x_1< x_2\)

\(\Rightarrow\left\{{}\begin{matrix}1.f\left(-2\right)\ge0\\\dfrac{S}{2}=\dfrac{4-m}{2}>-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4-2\left(m-4\right)-1\ge0\\\dfrac{8-m}{2}>0\end{matrix}\right.\) \(\Rightarrow m\le\dfrac{11}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=11\\b=2\end{matrix}\right.\) \(\Rightarrow a^2-b^3=121-8=113\)

\(I=\int e^xcosxdx\Rightarrow\left\{{}\begin{matrix}u=cosx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-sinx.dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=e^xcosx+\int e^xsinx.dx=e^xcosx+I_1\)

\(I_1=\int e^xsinx\Rightarrow\left\{{}\begin{matrix}u=sinx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=cosx.dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I_1=e^xsinx-\int e^xcosx.dx=e^x.sinx-I\)

\(\Rightarrow I=e^xcosx+e^xsinx-I\Rightarrow2I=e^x\left(cosx+sinx\right)\)

\(\Rightarrow I=e^x\left(\frac{1}{2}cosx+\frac{1}{2}sinx\right)+C\Rightarrow\left\{{}\begin{matrix}A=\frac{1}{2}\\B=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow A+B=1\)

I think that we have to prove \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

We have \(a+b+c=abc\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

We have \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=0\)( Because \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\))

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

So...

óc chó tự nghĩ đi nhá ahihihi