Hummmm

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 \(S=R/\left(-3;0\right)\)

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

Em 2k8 k biết làm có đúng k

ĐKXĐ : \(\left[{}\begin{matrix}x\le-1\\x\ge3\end{matrix}\right.\) 

Bpt \(\Leftrightarrow\left(x-2\right)\left[x+2-\sqrt{x^2-2x-3}\right]\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2;x+2\ge\sqrt{x^2-2x-3}\left(1\right)\\x\le2;x+2\le\sqrt{x^2-2x-3}\left(2\right)\end{matrix}\right.\)

(1) có :  \(x+2\ge\sqrt{x^2-2x-3}\Leftrightarrow\left(x+2\right)^2\ge x^2-2x-3\)

\(\Leftrightarrow6x+7\ge0\)  (Đ với \(x\ge2\) )

(2) có : \(\sqrt{x^2-2x-3}\ge x+2\)

TH1 : x + 2 < 0 <=> \(x< -2\)  ( Bpt luôn đúng ) 

TH2 : \(x+2\ge0\) ; Bp 2 vế rút gọn được : \(6x+7\le0\Leftrightarrow x\le\dfrac{-7}{6}\)

Khi đó : \(-2\le x\le\dfrac{-7}{6}\)

Vậy ... 

\(\sqrt[3]{2x^2+1}>\sqrt[3]{3x^2-1}\)

\(\Leftrightarrow2x^2+1>3x^2-1\)

\(\Leftrightarrow x^2< 2\)

\(\Leftrightarrow-\sqrt{2}< x< \sqrt{2}\)

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

a: ĐKXĐ: 2-|x-2|>=0

=>|x-2|<=2

=>-2<=x-2<=2

=>0<=x<=4

TH1: 0<=x<2

=>x-2<0

mà \(\sqrt{2-\left|x-2\right|}>=0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{2-\left|x-2\right|}\) >x-2 với mọi x thỏa mãn

=>NHận

=>0<=x<2

TH2: 2<=x<=4

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

=>\(2-\left|x-2\right|\ge\left(x-2\right)^2\)

=>\(\left(x-2\right)^2\le2-\left(x-2\right)=2-x+2=4-x\)

=>\(x^2-4x+4\le4-x\)

=>\(x^2-3x\le0\)

=>x(x-3)<=0

=>0<=x<=3

=>2<=x<=3

Vậy: 0<=x<=3

b: \(x^2+3x+2\ge2\cdot\sqrt{x^2+3x+5}\)

=>\(x^2+3x+5-2\sqrt{x^2+3x+5}-3\ge0\)

=>\(\left(\sqrt{x^2+3x+5}-3\right)\left(\sqrt{x^2+3x+5}+1\right)>=0\)

=>\(\sqrt{x^2+3x+5}-3\ge0\)

=>\(\sqrt{x^2+3x+5}\ge3\)

=>\(x^2+3x+5\ge9\)

=>\(x^2+3x-4\ge0\)

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

=>x>=1 hoặc x<=-4