`sqrt{x-2}-2>=sqrt{2x-5}-sqrt{x+1}`

`đk:x>=5/2`

`bpt<=>\sqrt{x-2}+\sqrt{x+1}>=\sqrt{2x-5}+2`

`<=>x-2+x+1+2\sqrt{(x-2)(x+1)}>=2x-5+4+4\sqrt{2x-5}`

`<=>2x-1+2\sqrt{(x-2)(x+1)}>=2x-1+4\sqrt{2x-5}`

`<=>2\sqrt{(x-2)(x+1)}>=4\sqrt{2x-5}`

`<=>sqrt{x^2-x-2}>=2sqrt{2x-5}`

`<=>x^2-x-2>=4(2x-5)`

`<=>x^2-x-2>=8x-20`

`<=>x^2-9x+18>=0`

`<=>(x-3)(x-6)>=0`

`<=>` \(\left[ \begin{array}{l}x \ge 6\\x \le 3\end{array} \right.\) 

Kết hợp đkxđ:

`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\) 

a,ĐK: x\(\ge\)1

⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)

⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

1.

ĐKXĐ: \(x=2\)

Xét \(x=2\), bất phương trình vô nghiệm

\(\Rightarrow\) bất phương trình đã cho vô nghiệm

\(\Rightarrow\) Không tồn tại \(a,b\) thỏa mãn

Đề bài lỗi chăng.

\(\sqrt{2x-1}\ge0\)

\(\Rightarrow BPT\ge0\) khi

\(3-2x-x^2\ge0\)

\(\Leftrightarrow x^2+2x-3\le0\)

\(\Leftrightarrow\left(x+1\right)^2-4\le0\)

\(\Leftrightarrow\left(x+1\right)^2\le4\)

\(\Leftrightarrow x+1\le2\)

\(\Rightarrow x\le1\)

a:

ĐKXĐ: x(x+3)>=0

=>x>=0 hoặc x<=-3

\(\left(x+5\right)\left(2-x\right)=3\cdot\sqrt{x^2+3x}\)

=>\(3\cdot\sqrt{x^2+3x}-\left(x+5\right)\left(2-x\right)=0\)

=>\(3\cdot\sqrt{x^2+3x}+\left(x+5\right)\left(x-2\right)=0\)

=>\(x^2+3x+3\cdot\sqrt{x^2+3x}-10=0\)

=>\(\left(\sqrt{x^2+3x}+5\right)\left(\sqrt{x^2+3x}-2\right)=0\)

=>\(\sqrt{x^2+3x}-2=0\)

=>\(\sqrt{x^2+3x}=2\)

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>x=-4(nhận) hoặc x=1(nhận)

e: \(\sqrt{2x^2+4x+1}=1-2x-x^2\)

=>\(\sqrt{2\left(x^2+2x\right)+1}=1-\left(2x+x^2\right)\)

=>\(2\left(x^2+2x\right)+1=\left\lbrack1-\left(2x+x^2\right)\right\rbrack^2=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1\) và \(1-2x-x^2\ge0\)

=>\(\left(x^2+2x\right)^2-4\left(x^2+2x\right)=0\) và \(x^2+2x-1\le0\)

=>\(\left(x^2+2x\right)\left(x^2+2x-4\right)=0\) và \(x^2+2x\le1\)

=>\(x^2+2x=0\)

=>x(x+2)=0

=>x=0 hoặc x=-2