a/ Với x = - 1 thì BĐT đúng.

Xét \(x\ne-1\)

Ta có: \(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}\le0\)

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

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

Đặt \(\frac{x}{\sqrt{x+1}}=t\)thì ta có bpt thành

\(t^3+3t^2-4\le0\)

\(\Leftrightarrow\left(t-1\right)\left(t+2\right)^2\le0\)

Tới đây thì đơn giản rồi b làm tiếp nhé.

Câu b còn lại mình nghĩ chỉ cần bình phương rồi chuyển cái chứa căn sang 1 bên không chứa căn sang 1 bên. Sau đó bình phương thêm 1 lần nữa rồi đặt nhân tử chung là ra :)

uk thank nha 

ĐKXĐ : \(1\le x\le3\)

Ta có \(\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}\ge x^3+10\)

<=> \(-2\sqrt{x-1}-2\sqrt{3-x}-8x\sqrt{2x}\le-2x^3-20\)

<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+2x^3-8x\sqrt{2x}+16\le0\)(1)

Đặt \(\sqrt{2x}=y\) => \(x=\dfrac{y^2}{2}\)

Khi đó \(2x^3-8x\sqrt{2x}+16=\dfrac{y^6}{4}-4y^3+16=\left(\dfrac{y^3-8}{2}\right)^2\)

Khi đó (1) <=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\le0\)(1)

mà \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\ge0\forall x;y\)(2) 

Từ (2)(1) => \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2=0\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{3-x}-1=0\\\dfrac{y^3-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\3-x=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\\\sqrt{2x}=2\end{matrix}\right.\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm bất phương trình

1.ĐK: \(x\ge\dfrac{1}{4}\)

bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)

\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)

\(\Leftrightarrow20x^2-x-1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)

2.ĐK: \(-2\le x\le\dfrac{5}{2}\)

bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)

\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)

\(\Leftrightarrow x^2< -x^2+x+6\)

\(\Leftrightarrow-2x^2+x+6>0\)

\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)

3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)

.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)

*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)

*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)