a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

\(\left(6x+7\right)^2.\left(3x+4\right).\left(x+1\right)=6\)

<=> \(\left(36x^2+84x+49\right)\left(3x^2+7x+4\right)=6\)

Đặt: \(3x^2+7x+4=t\)

=> \(36x^2+84x+49=12\left(3x^2+7x+4\right)+1=12t+1\)

Ta có phương trình ẩn t: 

\(t\left(12t+1\right)=6\)

<=> \(12t^2+t-6=0\)

<=> \(12t^2-8t+9t-6=0\)

<=> \(4t\left(3t-2\right)+3\left(3t-2\right)=0\)

<=> \(\left(4t+3\right)\left(3t-2\right)=0\)

<=> \(\orbr{\begin{cases}t=-\frac{3}{4}\\t=\frac{2}{3}\end{cases}}\)

Với \(t=-\frac{3}{4}\) ta có phương trình: \(3x^2+7x+4=-\frac{3}{4}\)

<=> \(x^2+\frac{7}{3}x+\frac{19}{12}=0\)

<=> \(x^2+2.x.\frac{7}{6}+\frac{49}{36}=-\frac{2}{9}\)

<=> \(\left(x+\frac{7}{6}\right)^2=-\frac{2}{9}\)phương trình vô nghiệm

+) Với \(t=\frac{2}{3}\)ta có: \(3x^2+7x+4=\frac{2}{3}\)

<=> \(x^2+\frac{7}{3}x+\frac{10}{9}=0\)

<=> \(x^2+2.x.\frac{7}{6}+\frac{49}{36}=\frac{1}{4}\)

<=> \(\left(x+\frac{7}{6}\right)^2=\frac{1}{4}\)

<=> \(x=-\frac{2}{3}\)

hoặc \(x=-\frac{5}{3}\)

Kết luận:...

Cách khác cô Chi nhé ! , nhưng cách này tới đấy xin cùy.

\(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6\)

\(108x^4+504x^3+879x^2+679x+196=6\)

\(108x^4+504x^3+879x^2+679x+190=0\)

Ta có: \(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}x\)

<=> \(\frac{7}{8}-\frac{7}{10}=\frac{2}{3}x+\frac{1}{3}x\)

<=>\(\frac{7}{40}=x\)

Vậy x=7/40

\(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}\)

\(\Leftrightarrow-\frac{1}{3}x-\frac{2}{3}x=\frac{7}{10}-\frac{7}{8}\)

\(\Leftrightarrow-x=-\frac{7}{40}\)

\(\Leftrightarrow x=\frac{7}{40}=0,175\)

Ta có: \(\frac{7}{8}-\frac{1}{3}.x=\frac{7}{10}+\frac{2}{3}.x\)

\(\Rightarrow\frac{7}{8}-\frac{7}{10}=\frac{2}{3}.x+\frac{1}{3}.x\)

\(\Rightarrow\frac{35}{40}-\frac{28}{40}=\left(\frac{2}{3}+\frac{1}{3}\right).x\)

\(\Rightarrow\frac{7}{40}=x\)

Vậy \(x=\frac{7}{40}\)

Chúc bạn hok tốt!