ĐKXĐ:

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

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

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

\(\Leftrightarrow x^2-3x+2=0\)

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

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)

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

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

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

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

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

ĐKXĐ: \(\begin{cases}3x^2-7x+3\ge0\\ x^2-3x+4\ge0\\ x^2-2\ge0\\ 3x^2-5x-1\ge0\end{cases}\)

=>\(\left[\begin{array}{l}x\le-\sqrt2\\ x\ge\frac{5+\sqrt{37}}{6}\end{array}\right.\)

BPT =>\(\sqrt{3x^2 - 7x + 3} - \sqrt{3x^2 - 5x - 1} > \sqrt{x^2 - 2} - \sqrt{x^2 - 3x + 4}\)

=>\(\dfrac{(3x^2 - 7x + 3) - (3x^2 - 5x - 1)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} > \dfrac{(x^2 - 2) - (x^2 - 3x + 4)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2x + 4}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}>\dfrac{3x - 6}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2(x - 2)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3(x - 2)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}>0\)

=>\((x-2)\left[\dfrac{-2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\right]>0\)

=>\((x - 2) \left[ \dfrac{2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} + \dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}} \right] < 0\)

=>x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: \(\left[\begin{array}{l}x\le-\sqrt2\\ \frac{5+\sqrt{37}}{6}\le x<2\end{array}\right.\)

Vậy: \(S = (-\infty, -\sqrt{2}] \cup \left[\dfrac{5+\sqrt{37}}{6}, 2\right)\)

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

=>\(\sqrt{3x+1}-2+2\sqrt{x+3}-4=3\sqrt{5x-1}-6\)

=>\(\dfrac{3x+1-4}{\sqrt{3x+1}+2}+2\left(\sqrt{x+3}-2\right)-3\left(\sqrt{5x-1}-2\right)=0\)

=>\(\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}+2\cdot\dfrac{x+3-4}{\sqrt{x+3}+2}-3\cdot\dfrac{5x-1-4}{\sqrt{5x-1}+2}=0\)

=>\(\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{2}{\sqrt{x+3}+2}-\dfrac{15}{\sqrt{5x-1}+2}\right)=0\)

=>x-1=0

=>x=1

ĐKXĐ: \(x>\dfrac{1}{5}\)

\(1-3x^2< \left(x+2\right)\sqrt[]{5x-1}+5x-1\)

\(\Leftrightarrow3x^2+5x-2+\left(x+2\right)\sqrt{5x-1}\ge0\)

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

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

\(\Leftrightarrow3x-1+\sqrt{5x-1}>0\)

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

TH1: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{5}\\1-3x< 0\end{matrix}\right.\) \(\Leftrightarrow x>\dfrac{1}{3}\)

TH2: \(\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\5x-1>9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\9x^2-11x+2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{9}< x\le\dfrac{1}{3}\)

Kết luận: \(x>\dfrac{2}{9}\)