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:

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

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

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)

\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)

\(< =>\dfrac{x-5}{4}=4x-17\)

\(< =>x-5=16x-68\)

\(< =>15x=68-5=63\)

\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)

b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)

\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)

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

\(< =>2x-2\sqrt{2x+1}-2=0\)

\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)

\(< =>\left(\sqrt{2x+1}-1\right)^2=4\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)

\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

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

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

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

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)

1.      \(2x^2-3x-5=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2,5\\x=-1\end{cases}}\)

Vậy tập ngiệm của phương trình là \(S=\left\{2,5;-1\right\}\)

2x²-3x-5=0

2x²+2x-5x-5=0

2x(x+1)+5(x+1)=0

(x+1)(2x+5)=0

TH1 x+1=0 <=>x=-1

TH2 2x+5=0<=>2x=-5<=>x=-5/2

2. ta có:

2(x-2y)-(2x+y)=-1.2-8

2x-4y-2x-y=-2-8

-5y=-10

y=2

thay vào 

x-2y=-1 ( với y=2)

<=> x-2.2=-1

x-4=-1

x=3