ĐKXĐ : \(x\ge-3\)

+ Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+3}\ge0\\b=\sqrt[3]{x}\end{matrix}\right.\) ta có :

\(\left\{{}\begin{matrix}a+b=3\\a^2-b^3=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3-b\\\left(3-b\right)^2-b^3=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3-b\\b^3-b^2+6b-6=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=3-b\\\left(b^2+6\right)\left(b-1\right)=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{a+3}=2\\\sqrt[3]{x}=1\end{matrix}\right.\Rightarrow x=1\) ( TM )

2. \(pt\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\Rightarrow3x-5>0\Rightarrow x>\frac{5}{3}\)

+ \(pt\Leftrightarrow\left(\sqrt{x^2+12}-4\right)-\left(\sqrt{x^2+5}-3\right)-\left(3x-6\right)=0\)

\(\Leftrightarrow\frac{x^2-4}{\sqrt{x^2+12}+4}-\frac{x^2-4}{\sqrt{x^2+5}+3}-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{\sqrt{x^2+12}+4}-\frac{x+2}{\sqrt{x^2+5}+3}-3\right)=0\) (1)

+ \(\forall x>\frac{5}{3}\) ta có: \(\left\{{}\begin{matrix}x+2>0\\\sqrt{x^2+12}+4>\sqrt{x^2+5}+3\end{matrix}\right.\)

\(\Rightarrow\frac{x+2}{\sqrt{x^2+12}+4}< \frac{x+2}{\sqrt{x^2+5}+3}\Rightarrow\frac{x+2}{\sqrt{x^2+12}+4}-\frac{x+2}{\sqrt{x^2+5}+3}-3< 0\) nên từ (1) suy ra:

\(x-2=0\Leftrightarrow x=2\) ( TM )

Áp dụng bđt Cauchy hết nha

a) \(A=\sqrt{\left(a+1\right)\cdot\frac{3}{2}}\le\frac{a+1+\frac{3}{2}}{2}=\frac{a+\frac{5}{2}}{2}\). Dấu "=" \(\Leftrightarrow a+1=\frac{3}{2}\Leftrightarrow a=\frac{1}{2}\)

+ Tương tự : \(\sqrt{\left(b+1\right)\cdot\frac{3}{2}}\le\frac{b+\frac{5}{2}}{2}\) Dấu "=" \(\Leftrightarrow b=\frac{1}{2}\)

Do đó : \(\sqrt{\frac{3}{2}}\cdot A\le\frac{a+b+5}{2}=3\) \(\Rightarrow A\le\sqrt{6}\)

Dấu "=" \(\Leftrightarrow a=b=\frac{1}{2}\)

b) \(B=x\cdot x\left(1-2x\right)\le\left(\frac{x+x+1-2x}{3}\right)^3=\frac{1}{27}\)

Dấu "=" \(\Leftrightarrow x=1-2x\Leftrightarrow x=\frac{1}{3}\)

c) \(C=\frac{1}{2}\left(2x+2\right)\left(1-2x\right)\le\frac{1}{2}\left(\frac{2x+2+1-2x}{2}\right)^2=\frac{9}{8}\)

Dấu "=" \(\Leftrightarrow2x+2=1-2x\Leftrightarrow x=-\frac{1}{4}\)

5. Tìm min???

Áp dụng bđt Cauchy ta có:

\(a^3+b^3+1\ge3\sqrt[3]{a^3\cdot b^3\cdot1}=3ab\) . Dấu "=" \(\Leftrightarrow a=b=1\)

+ Tương tự : \(a^3+c^3+1\ge3ac\). Dấu "=" \(\Leftrightarrow a=c=1\)

\(b^3+c^3+1\ge3bc\). Dấu "=" \(\Leftrightarrow b=c=1\)

Do đó : \(2D+3\ge3\left(ab+bc+ca\right)\Rightarrow D\ge3\)

Min \(D=3\Leftrightarrow a=b=c=1\)

3. a) \(A=x+\frac{1}{x-1}=x-1+\frac{1}{x-1}+1\ge2\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+1=3\)

Dấu "=" \(\Leftrightarrow x-1=\frac{1}{x-1}\Leftrightarrow x=2\)

Min \(A=3\Leftrightarrow x=2\)

b) \(B=\frac{4}{x}+\frac{1}{4y}=\frac{4}{x}+4x+\frac{1}{4y}+4y\cdot-4\left(x+y\right)\)

\(\ge2\sqrt{\frac{4}{x}\cdot4x}+2\sqrt{\frac{1}{4y}\cdot4y}-4\cdot\frac{5}{4}=5\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x}=4x\\\frac{1}{4y}=4y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)

Min \(B=5\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)

4. Chắc đề là tìm min???

\(C=a+b+\frac{1}{a}+\frac{1}{b}\ge a+b+\frac{4}{a+b}=a+b+\frac{1}{a+b}+\frac{3}{a+b}\)

\(\ge2\sqrt{\left(a+b\right)\cdot\frac{1}{a+b}}+\frac{3}{1}=5\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a+b=\frac{1}{a+b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\frac{1}{2}\)

Min \(C=5\Leftrightarrow a=b=\frac{1}{2}\)

1. Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:

\(\left(\frac{1}{p-a}+\frac{1}{p-b}\right)+\left(\frac{1}{p-b}+\frac{1}{p-c}\right)+\left(\frac{1}{p-c}+\frac{1}{p-a}\right)\)

\(\ge\frac{4}{2p-a-b}+\frac{4}{2p-b-c}+\frac{4}{2p-a-c}\) \(=\frac{4}{c}+\frac{4}{a}+\frac{4}{b}\)

\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" \(\Leftrightarrow a=b=c\)

2. Áp dụng bđt Cauchy ta có :

\(a\sqrt{b-1}=a\sqrt{\left(b-1\right)\cdot1}\le a\cdot\frac{b-1+1}{2}=\frac{ab}{2}\) . Dấu "=" \(\Leftrightarrow b-1=1\Leftrightarrow b=2\)

+ Tương tự : \(b\sqrt{a-1}\le\frac{ab}{2}\). Dấu "=" \(\Leftrightarrow a=2\)

Do đó: \(a\sqrt{b-1}+b\sqrt{a-1}\le ab\). Dấu "=" \(\Leftrightarrow a=b=2\)

\(y=\frac{2x^3-2x^2+1}{x^2}=2x-2+\frac{1}{x^2}=x+x+\frac{1}{x^2}-2\)

\(\ge3\sqrt[3]{x\cdot x\cdot\frac{1}{x^2}}-2=1\)

Dấu "=" \(\Leftrightarrow x=\frac{1}{x^2}\Leftrightarrow x=1\)

\(D=\left(2x+1\right)\left(4-x\right)\left(4-x\right)\le\left(\frac{2x+1+4-x+4-x}{3}\right)^3=27\)

Dấu "=" \(\Leftrightarrow2x+1=4-x\Leftrightarrow x=1\)

\(2N=4x^2+4xy+10y^2-16x-44y+4038\)

\(=4x^2+4x\left(y-4\right)+\left(y-4\right)^2-\left(y-4\right)^2+10y^2-44y+4038\)

\(=\left(2x+y-4\right)^2+9y^2-36y^2+36+3986\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2+3986\ge3986\forall x,y\)

\(\Rightarrow N\ge1993\forall x,y\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y-4\right)^2=0\\\left(3y-6\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

+ Áp dụng bđt Cauchy ta có:

\(E=\frac{1}{4}\cdot2x\cdot2x\left(9-4x\right)\le\frac{1}{4}\left(\frac{2x+2x+9-4x}{3}\right)^3=\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow2x=9-4x\Leftrightarrow x=\frac{3}{2}\)