1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

a: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x^2-x-20\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}\left(x+1\right)\left(5x+9\right)\ge0\\ \left(x-5\right)\left(x+4\right)\ge0\\ x\ge-1\end{cases}\)

=>x>=5

TA có: \(\sqrt{5x^2+14x+9} \le 5\sqrt{x+1} + \sqrt{x^2-x-20}\)

=>\(5x^2+14x+9 \le 25(x+1) + x^2-x-20 + 10\sqrt{(x+1)(x^2-x-20)}\)

=>\(5x^2+14x+9 \le x^2 + 24x + 5 + 10\sqrt{(x+1)^2(x-5)}\)

=>\(4x^2 - 10x + 4 \le 10(x+1)\sqrt{x-5}\)

=>\(2x^2 - 5x + 2 \le 5(x+1)\sqrt{x-5}\)

=>\((2x-1)(x-2) \le 5(x+1)\sqrt{x-5}\) (1)

Đặt \(t=\sqrt{x-5}\ge0\implies x=t^2+5\)

(1) sẽ trở thành: \(2(t^2+5)^2 - 5(t^2+5) + 2 \le 5(t^2+6)t\)

=>\(2(t^4 + 10t^2 + 25) - 5t^2 - 25 + 2 \le 5t^3 + 30t\)

=>\(2t^4 + 20t^2 + 50 - 5t^2 - 23 \le 5t^3 + 30t\)

=>\(2t^4 - 5t^3 + 15t^2 - 30t + 27 \le 0\)

=>\((t-1)(2t-3)(t^2 + 6) \le 0\)

=>(t-1)(2t-3)<=0

=>1<=t<=3/2

=>\(1\le\sqrt{x-5}\le\frac{3}{2}\)

=>\(1\le x-5\le\frac{9}{4}\)

\(\iff6\le x\le\frac{29}{4}\)

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...

ĐKXĐ: \(x\le2\)

Xét trên miền xác định:

\(\Leftrightarrow\dfrac{2x^3+3x}{7-2x}-1+1-\sqrt{2-x}>0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2x^2+2x+7\right)}{7-2x}+\dfrac{x-1}{1+\sqrt{2-x}}>0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{2x^2+2x+7}{7-2x}+\dfrac{1}{1+\sqrt{2-x}}\right)>0\)

\(\Leftrightarrow1< x\le2\)

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

=>\(-2\sqrt{3x}=-4\)

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

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

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

a: ĐKXĐ: x<0 hoặc x>1

TH1: x<0

=>x-1<-1<0

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

\(\frac{x - 1}{(x - 1) - \sqrt{-x(1-x)}} = \frac{-(1-x)}{-(1-x) - \sqrt{-x}\sqrt{1-x}} = \frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}}\)

Bất phương trình sẽ trở thành: \(\frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}} \ge 2x\)

x<0

=>2x<0

mà \(\frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}}>0\forall x\) thỏa mãn

nên Bất phương trình thỏa mãn với mọi x<0

TH2: x>1

=>x-1>0 và \(\sqrt{x^2 - x} = \sqrt{x(x-1)} = \sqrt{x}\sqrt{x-1}\)

Bất phương trình sẽ tương đương: \(\frac{\sqrt{x-1}}{\sqrt{x-1} - \sqrt{x}} \ge 2x\)

mà \(\frac{\sqrt{x-1}}{\sqrt{x-1} - \sqrt{x}}=\frac{\sqrt{x-1}(\sqrt{x-1} + \sqrt{x})}{(\sqrt{x-1} - \sqrt{x})(\sqrt{x-1} + \sqrt{x})}=\frac{\sqrt{x-1}(\sqrt{x-1} + \sqrt{x})}{(x-1) - x}=-\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})\)

nên \(-\sqrt{x-1}(\sqrt{x-1} + \sqrt{x}) \ge 2x\)

=>\(\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})+2x\le0\) (vô lý vì \(\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})+2x>0\) với mọi x>1)

=>x∈∅

Vậy: x<0

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)