1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)

2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)

\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)

mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)

Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)

\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)

\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

=>\(\sqrt{\left(x+3\right)\left(x-7\right)}=x+4\)

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

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

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

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\) (1)

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

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

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

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

=>\(\sqrt{x-1}<1\)

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{3x+1}-\sqrt{2x+3}=\dfrac{x-2}{4}\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{x-2}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt{3x+1}+\sqrt{2x+3}}=\dfrac{1}{4}\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow\sqrt{3x+1}+\sqrt{2x+3}=4\)

\(\Leftrightarrow5x+4+2\sqrt{6x^2+11x+3}=16\)

\(\Leftrightarrow2\sqrt{6x^2+11x+3}=12-5x\) (\(x\le\dfrac{12}{5}\))

\(\Leftrightarrow4\left(6x^2+11x+3\right)=\left(12-5x\right)^2\)

\(\Leftrightarrow x^2-164x+132=0\Rightarrow\left[{}\begin{matrix}x=82-8\sqrt{103}\\x=82+8\sqrt{103}>\dfrac{12}{5}\left(loại\right)\end{matrix}\right.\)

1: ĐKXĐ: x>=8/3

\(\sqrt{3x-8}-\sqrt{x+1}=\frac{2x-11}{5}\)

=>\(\sqrt{3x-8}-1+2-\sqrt{x+1}=\frac{2x-11}{5}+1\)

=>\(\frac{3x-8-1}{\sqrt{3x-8}+1}+\frac{4-x-1}{2+\sqrt{x+1}}=\frac{2x-11+5}{5}\)

=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-8}+1}-\frac{1}{2+\sqrt{x+1}}-\frac25\right)=0\)

=>x-3=0

=>x=3(nhận)

3: ĐKXĐ: -5/2<=x<=5/2

Đặt \(a=\sqrt{5+2x};b=\sqrt{5-2x}\)

=>\(ab=\sqrt{\left(5+2x\right)\left(5-2x\right)}=\sqrt{25-4x^2}\)

Theo đề, ta có: a+b+5=3ab

=>3ab-a-b-5=0

=>a(3b-1)-b+1/3-16/3=0

=>\(3a\left(b-\frac13\right)-\left(b-\frac13\right)=\frac{16}{3}\)

=>\(\left(b-\frac13\right)\left(3a-1\right)=\frac{16}{3}\)

=>(3a-1)(3b-1)=16

=>(3a-1;3b-1)∈{(1;16);(16;1);(2;8);(8;2);(4;4)}

=>(3a;3b)∈{(2;17);(17;2);(3;9);(9;3);(5;5)}

=>(a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1);(5/3;5/3)}

mà a<>b

nên (a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1)}

TH1: a=2/3 và b=17/3

=>\(\begin{cases}5+2x=\frac49\\ 5-2x=\frac{289}{9}\end{cases}\Rightarrow\begin{cases}2x=\frac49-5=\frac49-\frac{45}{9}=-\frac{41}{9}\\ 2x=5-\frac{289}{9}=-\frac{244}{9}\end{cases}\)

=>x∈∅

TH2: a=17/3 và b=2/3

=>\(\begin{cases}5+2x=\frac{289}{9}\\ 5-2x=\frac49\end{cases}\Rightarrow\begin{cases}2x=\frac{289}{9}-5=\frac{244}{9}\\ 2x=5-\frac49=\frac{41}{9}\end{cases}\)

=>x∈∅

TH3: a=1 và b=3

=>5+2x=1 và 5-2x=9

=>2x=-4 và 2x=5-9=-4

=>x=-2(nhận)

TH4: a=3 và b=1

=>5+2x=9 và 5-2x=1

=>2x=4 và 2x=4

=>x=2(nhận)

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

\(\Leftrightarrow2\left(3x+1\right)\sqrt{2x^2-1}=10x^2+3x-6\)

Đặt \(t=\sqrt{2x^2-1}\left(t\ge0\right)\)  \(\left(1\right)\) nên ta có phương trình:

\(4t^2-2\left(3x+1\right)t+2x^2+3x-2=0\)

Ta có: \(\Delta'=\left(3x+1\right)^2-4\left(2x^2+3x-2\right)=\left(x-3\right)^2\)

⇒ Phương trình có hai nghiệm phân biệt

\(t_1=\dfrac{2x-1}{2}\)

\(t_2=\dfrac{x+2}{2}\)

Thay lần lượt các giá trị của t vào (1) nên: \(x\in\left\{\dfrac{-1+\sqrt{6}}{2};\dfrac{2+\sqrt{60}}{7}\right\}\)

a. ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\-1\le x< 0\end{matrix}\right.\)

Do \(x\ne0\) nên pt tương đương:

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

\(\Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\dfrac{1}{x}}=t\ge0\)

\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x-\dfrac{1}{x}=1\)

\(\Rightarrow x^2-x-1=0\Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}\)

b.

ĐKXĐ: \(x\ge0\)

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

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

\(\Leftrightarrow x-1+\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-1}{\sqrt{x+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x+3}+2}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x+3}+1}{\sqrt{x+3}+2}\right)=0\)

\(\Leftrightarrow x-1=0\)

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)