\(A^2=9-x+2\sqrt{\left(2x+5\right)\left(4-3x\right)}\ge9-x\ge9-\frac{4}{3}=\frac{23}{3}\)

\(\Rightarrow A\ge\sqrt{\frac{23}{3}}\Rightarrow a+b=26\)

ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\Leftrightarrow x^3-1+2x-1-\sqrt{3x-2}+x+1-\sqrt{x+3}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{4x^2-7x+3}{2x-1+\sqrt{3x-2}}+\frac{x^2+x-2}{x+1+\sqrt{x+3}}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\frac{\left(x-1\right)\left(4x-3\right)}{2x-1+\sqrt{3x-2}}+\frac{\left(x-1\right)\left(x+2\right)}{x+1+\sqrt{x+3}}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1+\frac{4x-3}{2x-1+\sqrt{3x-2}}+\frac{x+2}{x+1+\sqrt{x+3}}\right)\le0\)

\(\Leftrightarrow x-1\le0\) (ngoặc đằng sau luôn dương)

\(\Rightarrow x\le1\Rightarrow\frac{2}{3}\le x\le1\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=1\end{matrix}\right.\) \(\Rightarrow a+b=5\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{12-x}=a\\\sqrt[3]{4+x}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a+b=2\\a^3+b^3=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=2\\\left(a+b\right)\left(a^2+b^2-ab\right)=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=2-a\\a^2+b^2-ab=8\end{matrix}\right.\)

\(\Rightarrow a^2+\left(2-a\right)^2-a\left(2-a\right)-8=0\)

\(\Leftrightarrow3a^2-6a-4=0\Rightarrow a=\frac{3\pm\sqrt{21}}{2}\)

\(\Rightarrow\sqrt[3]{12-x}=\frac{3\pm\sqrt{21}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{36-16\sqrt{21}}{9}\\x=\frac{36+16\sqrt{21}}{9}\end{matrix}\right.\)

Bài toán có tới 2 nghiệm thỏa mãn? b có 2 giá trị là \(\pm16\) lấy cái nào?

\(\sqrt{x^2+4x+3m+1}=x+3\)

\(\Leftrightarrow x^2+4x+3m+1=\left(x+3\right)^2\)

\(\Leftrightarrow x^2+4x+3m+1=x^2+6x+9\)

\(\Leftrightarrow2x=3m-8\)

\(\Leftrightarrow x=\frac{3m-8}{2}\)

Với x=\(\frac{3m-8}{2}\Rightarrow\left(\frac{3m-8}{2}\right)^2+4\cdot\frac{3m-8}{2}+3m+1\ge0\)

\(\Leftrightarrow\frac{9m^2-48m+64}{4}+6m-16+3m+1\ge0\)

\(\Leftrightarrow9m^2-12m+4\ge0\)

\(\Leftrightarrow\left(3m-2\right)^2\ge0\)(luôn đúng)

Dấu "=" xảy ra <=> \(3m-2=0\Leftrightarrow m=\frac{2}{3}\)

\(\Rightarrow a=2;b=3\)

\(\Rightarrow4a^2+3b^2+7=4\cdot2^2+3\cdot3^2+7=50\)

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 "=" \(\Leftrightarrow a=b=2\)