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:

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

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

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

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

\(\Rightarrow t=2\)

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

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

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

\(\Leftrightarrow...\)

Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:

\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)

\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)

Ta có:

\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)

\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)

Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)

Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

ĐK: \(x\ge2\)

\(\dfrac{\sqrt{x^2+1}-\sqrt{x+1}}{x^2+\sqrt{3x-6}}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+1}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x+1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x^2+1\ge x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1\le x\le0\\x\ge1\end{matrix}\right.\)

Kết hợp điều kiện xác định ta được \(x\ge2\)

Lag tí -.-'

`ĐK:2<=x<=6`

BP 2 vế ta có:

`x-2+6-x+2\sqrt{(x-2)(6-x)}=x^2-8x+24`

`<=>4+2\sqrt{(x-2)(6-x)}=x^2-8x+24`

`<=>2\sqrt{(x-2)(6-x)}=x^2-8x+20`

`<=>2sqrt{-x^2+8x-12}=x^2-8x+20`

`<=>-x^2+8x-20+2sqrt{-x^2+8x-12}=0`

`<=>-x^2+8x-12+2sqrt{-x^2+8x-12}-8=0`

Đặt `sqrt{-x^2+8x-12}=a(a>=0)`

`pt<=>a^2+2a-8=0`

`<=>a=2(tm),a=-4(l)`

`<=>-x^2+8x-12=4`

`<=>x^2-8x+16=0`

`<=>(x-4)^2=0<=>x=4(tmđk)`

Vậy `S={4}`

\(\sqrt{x-2}+\sqrt{6-x}\text{=}\sqrt{x^2-8x+24}\)

\(ĐKXĐ:2\le x\le6\)

Xét VP của pt ta thấy : \(\sqrt{x^2-8x+24}\text{=}\sqrt{x^2-8x+16+8}\)

\(\text{=}\sqrt{\left(x-4\right)^2+8}\)

\(\Rightarrow VP\ge\sqrt{8}\)

Xét VT của pt ta có :

\(VT^2\text{=}x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

\(VT^2\text{=}4+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Áp dụng BĐT cô si cho 2 số không âm ta có :

\(2\sqrt{\left(x-2\right)\left(6-x\right)}\le\left(\sqrt{x-2}\right)^2+\left(\sqrt{6-x}\right)^2\)

\(\text{=}x-2+6-x\text{=}4\)

\(\Rightarrow VT^2\le8\)

\(\Rightarrow VT\le\sqrt{8}\)

Để \(VT\text{=}VP\) \(\Leftrightarrow\left\{{}\begin{matrix}x-4\text{=}0\\\sqrt{x-2}\text{=}\sqrt{6-x}\end{matrix}\right.\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy...........

\(\sqrt{x+2}-\sqrt{x-6}>2\)(ĐK: x\(\ge\)6)

\(\Leftrightarrow\sqrt{x+2}>2+\sqrt{x-6}\)

\(\Leftrightarrow x+2>4+4\sqrt{x-6}+x-6\)

\(\Leftrightarrow4>4\sqrt{x-6}\)

<=>\(1>\sqrt{x-6}\)

<=>1>x-6

<=>x<7 Mà x\(\ge\)6 =>x=6

6   

x=6