`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:

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

c) Đặt \(t=\sqrt{\left(x-3\right)\left(8-x\right)}\left(t\ge0\right)=\sqrt{-x^2+11x-24}\Rightarrow t^2-2=-x^2+11x-26\)

\(\left(1\right)\Rightarrow t\ge t^2-2\Leftrightarrow t^2-t-2\le0\Leftrightarrow-1\le t\le2\Rightarrow0\le t\le2\Rightarrow0\le-x^2+11x-24\le4\Leftrightarrow\left\{{}\begin{matrix}3\le x\le8\\\left[{}\begin{matrix}x\le4\\x\ge7\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\le x\le4\\7\le x\le8\end{matrix}\right.\)

Vậy tập nghiệm của bpt là \([3;4]\cup[7;8]\)

Bn tk hen:

ĐKXĐ: \(x>0\)

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

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

\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{\sqrt{x^2-x+1}+\sqrt{x}}.\sqrt{\dfrac{x^2+x+1}{x^2+1}}+\dfrac{\left(x-1\right)^2}{x}\ge0\) (luôn đúng \(\forall x>0\))

Vậy nghiệm của BPT đã cho là \(x>0\)

e: \(\begin{cases}x\left(x+5\right)<4x+2\\ \left(2x-1\right)\left(x+3\right)\ge4x\end{cases}\Rightarrow\begin{cases}x^2+5x-4x-2<0\\ 2x^2+6x-x-3-4x\ge0\end{cases}\)

=>\(\begin{cases}x^2+x-2<0\\ 2x^2+x-3\ge0\end{cases}\Rightarrow\begin{cases}\left(x+2\right)\left(x-1\right)<0\\ 2x^2+3x-2x-3\ge0\end{cases}\)

=>\(\begin{cases}-2

=>-2<x<=-1

f: ĐKXĐ: x∉{1;4;2;5}

Ta có: \(\frac{1}{x^2-5x+4}\le\frac{1}{x^2-7x+10}\)

=>\(\frac{1}{x^2-5x+4}-\frac{1}{x^2-7x+10}\le0\)

=>\(\frac{x^2-7x+10-x^2+5x-4}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)

=>\(\frac{-2x+6}{\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-5\right)}\le0\)

=>\(\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\ge0\)

Đặt \(A=\frac{x-3}{\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)}\)

Đặt x-3=0

=>x=3

Đặt x-1=0

=>x=1

Đặt x-2=0

=>x=2

Đặt x-4=0

=>x=4

Đặt x-5=0

=>x=5

Bảng xét dấu:

Theo bãng xét dấu, ta có: A>=0 khi 1<x<2; 3<=x<4; x>5

bài 2

ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)

\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)

Áp dụng bất đẳng thức Bunhiacopxki ta có;

\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)

\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)

\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)

Dấu \(=\)xảy ra khi \(a=b=c=1\)

câu 1 dễ mà liên hợp đi x=\(\frac{4}{5}\)