a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)

\(Xét-mẫu-của-biểu-thức:\left(đk:x\ge1\right).ta-có:x-\sqrt{2\left(x^2+5\right)}=\frac{-\left(x^2+10\right)}{x+\sqrt{2\left(x^2+5\right)}}< 0\\ .\)Vậy nó luôn <0 với đk x>=1
\(Xét-tử:đặt-nó-bằng-A=\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)=2\sqrt{x-1}\left(2x-1\right)-\left(x-1\right)\left(x+4\right)\\ =\sqrt{x-1}\left(2\left(2x-1\right)-\sqrt{x-1\left(x+4\right)}\right)\ge0.\\ \)\(=>\left(2\left(2x-1\right)-\sqrt{\left(x-1\right)}\left(x+4\right)\right)\ge0< =>\frac{\left(5-x\right)\left(x-2\right)^2}{2\left(2x-1\right)+\left(x-1\right)\left(x+4\right)}\ge0< =>x\le5\) Vậy . \(1\le x\le5\)
 

Thank you ^^^

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

Điều kiện \(x^2-2x\ge0\Leftrightarrow\left[\begin{array}{nghiempt}x\ge2\\x\le0\end{array}\right.\) khi đó :

Bất phương trình \(\Leftrightarrow3^{\sqrt{x^2-2x}}\ge\left(3\right)^{\sqrt{\left(x-1\right)^2}-x}\Leftrightarrow\sqrt{x^2-2x}\ge\left|x-1\right|-x\)

- Khi \(x\ge2\Rightarrow x-1>0\) nên bất phương trình \(\sqrt{x^2-2x}\ge-1\) đúng với mọi \(x\ge2\)

- Khi \(x\le0\Rightarrow x-1< 0\) nên bất phương trình \(\sqrt{x^2-2x}\ge1-2x\)

                                                                 \(\Leftrightarrow\begin{cases}x^2-2x\ge1-4x+4x^2\\x\le0\end{cases}\) vô nghiệm

Vậy tập nghiệm của bất phương trình là : S = [2;\(+\infty\) )