1. Đợi chút t tìm cách ngắn gọn.

2. ĐK: \(\left\{{}\begin{matrix}2x^2+8x+6\ge0\\x^2-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge1\\x=-1\end{matrix}\right.\) (*)

BPT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2+8x+5+2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\le\left(2x+2\right)^2\left(1\right)\end{matrix}\right.\)

Giải (1) \(\Leftrightarrow x^2-1-2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\right)\ge0\)

TH1: \(\sqrt{x^2-1}=0\Leftrightarrow x=\pm1\) (tm)

TH2: \(x^2-1\ne0\)

\(\Leftrightarrow\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\ge2\sqrt{2x^2+8x+6}\)

\(\Leftrightarrow x^2-1\ge8x^2+32x+24\)

\(\Leftrightarrow7x^2+32x+25\le0\)

\(\Leftrightarrow-\frac{25}{7}\le x\le-1\) kết hợp đk (*) và đk để giải bpt

=>\(x=-1\)

Vậy \(x=\pm1\)

3. ĐK: \(x\ge\frac{4}{5}\)

\(BPT\Leftrightarrow\sqrt{5x-4}-\sqrt{3x-2}+\sqrt{4x-3}-\sqrt{2x-1}>0\)

\(\Leftrightarrow\frac{2x-2}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{2x-2}{\sqrt{4x-3}+\sqrt{2x-1}}>0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{1}{\sqrt{4x-3}+\sqrt{2x-1}}\right)>0\)

\(\Leftrightarrow x-1>0\) \(\Leftrightarrow x>1\)

Vậy \(x>1\)

ĐKXĐ: \(x\ge3\)

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

\(\Leftrightarrow x-3< x-1+x-2+2\sqrt{\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(x-2\right)}>-x\)

Do \(x\ge3\Rightarrow-x< 0\) , mà \(\sqrt{\left(x-1\right)\left(x-2\right)}>0\Rightarrow\) BPT đúng với mọi \(x\ge3\)

Vậy nghiệm của BPT là \(x\ge3\)

Hoặc có thể biện luận dễ dàng như sau:

\(\sqrt{x-3}< \sqrt{x-1}\) \(\forall x\ge3\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP>0\end{matrix}\right.\)

\(\Rightarrow\) BPT đúng với mọi \(x\ge3\)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)

b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)

c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)

\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)

d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)

\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)

a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)

<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)

<=> \(\sqrt{x}+8=28\)

<=> \(\sqrt{x}=28-8\)

<=> \(\sqrt{x}=20\)

<=> \(\left(\sqrt{x}\right)^2=20^2\)

<=> x = 400

=> x = 400

b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)

<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)

<=> \(3\sqrt{x}+5=\sqrt{x}+12\)

<=> \(3\sqrt{x}=\sqrt{x}+12-5\)

<=> \(3\sqrt{x}=\sqrt{x}+7\)

<=> \(3\sqrt{x}-\sqrt{x}=7\)

<=> \(2\sqrt{x}=7\)

<=> \(\sqrt{x}=\frac{7}{2}\)

<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)

<=> \(x=\frac{49}{4}\)

=> \(x=\frac{49}{4}\)

c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)

<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)

<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)

<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)

<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)

<=> \(8\sqrt{x}=6\sqrt{x}+4\)

<=> \(8\sqrt{x}-6\sqrt{x}=4\)

<=> \(2\sqrt{x}=4\)

<=> \(\sqrt{x}=2\)

<=> \(\left(\sqrt{x}\right)^2=2^2\)

<=> x = 4

=> x = 4

d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)

<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)

<=>\(2\sqrt{3x}=6\sqrt{3x}\)

<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)

<=>\(-4\sqrt{3x}=0\)

<=> \(\sqrt{3x}=0\)

<=> \(\left(\sqrt{3x}\right)^2=0^2\)

<=> 3x = 0

<=> x = 0

=> x = 0