\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

TA CÓ:

\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)

\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)

\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)

\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

a.\(\left(x^2+2x+5\right)\left(x^2+4x\right)=0\)

Ta có: \(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0;\forall x\)

 \(\Rightarrow x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

b.\(\left(x^2-4x+4\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)

c.\(1,2x^3-x^2-0,2x=0\)

\(\Leftrightarrow x\left(1,2x^2-x-0,2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{1}{6}\end{matrix}\right.\)

1, x(x-1)=2(x-1)

<=> x(x-1)-2(x-1)=0

<=> (x-2)(x-1)=0

<=>x=2 hoặc x=1 

vậy ...

2, (x+2)(2x-3)=x^2 -4

<=>(x+2)(2x-3)=(x-2)(x+2)

<=> (x+2)(2x-3)-(x-2)(x+2)=0

<=> (x+2)(2x-3-x+2)=0 

<=> x=-2 hoặc x=1

vây... 

3,x^2 +3x +2=0 

<=> x^2 +x+2x+2=0 

<=>(x+2)(x+1)=0

<=> x=-2 hoặc x=-1 

vậy ...

5, x^3+x^2-12x =0

<=> x(x^2+x-12)=0

<=>x(x^2-3x+4x-12)=0

<=>x(x+4)(x-3)=0 

<=> x=0 hoặc x=-4 hoặc x=3

vậy ... 

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

Rảnh rỗi thật sự .-.

cả 2 pt đều giải theo kiểu cái đầu nhóm với cái cuối, 2 cái ở giữa nhóm với nhau. sau đó giải theo cách đặt ẩn phụ

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)

\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

Đặt \(x^2+7x=a\), nên ta có :

\(\left(a+10\right)\left(a+12\right)-24=0\)

\(\Leftrightarrow\left(x+11-1\right)\left(x+11+1\right)-24=0\)

\(\Leftrightarrow\left[\left(x+11\right)^2-1\right]-24=0\)

\(\Leftrightarrow\left(x+11\right)^2-25=0\)

\(\Leftrightarrow\left(x+11-5\right)\left(x+11+5\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+16\right)=0\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-16\end{cases}}\)

Đặt \(u=x^2-x\)

Phương trình trở thành \(u^2-4u+4=0\)

\(\Leftrightarrow\left(u-2\right)^2=0\)

\(\Leftrightarrow u-2=0\)

\(\Rightarrow x^2-x=2\)

\(\Rightarrow x^2-x-2=0\)

Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1+3}{2}=2\\x=\frac{1-3}{2}=-1\end{cases}}\)

Đặt \(2x+1=w\)

Phương trình trở thành \(w^2-w=2\)

\(\Rightarrow\orbr{\begin{cases}w=2\\w=-1\end{cases}}\)

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

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

Đặt\(x^2-3x=k\)

Phương trình trở thành \(k^2+5k+6=0\)

\(\Leftrightarrow\left(k+2\right)\left(k+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}k=-2\\k=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x=-2\\x^2-3x=-3\end{cases}}\)

Vậy \(x\in\left\{1;2\right\}\)