1: 5(2-3x)(x-2)=3(1-3x)

=>5(3x-2)(x-2)=3(3x-1)

=>\(5\cdot\left(3x^2-6x-2x+4\right)=9x-3\)

=>\(15x^2-40x+20-9x+3=0\)

=>\(15x^2-49x+23=0\)

\(\Delta=49^2-4\cdot15\cdot23=1021>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{49-\sqrt{1021}}{2\cdot15}=\frac{49-\sqrt{1021}}{30}\\ x=\frac{49+\sqrt{1021}}{2\cdot15}=\frac{49+\sqrt{1021}}{30}\end{array}\right.\)

2: \(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\)

3: \(4x^2-9=0\)

=>\(4x^2=9\)

=>\(x^2=\frac94\)

=>\(\left[\begin{array}{l}x=\frac32\\ x=-\frac32\end{array}\right.\)

4: \(5x^2-10x=0\)

=>5x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

5: \(x^2-3x=-2\)

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

6: |x-5|-3=0

=>|x-5|=3

=>\(\left[\begin{array}{l}x-5=3\\ x-5=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=2\end{array}\right.\)

a: (3x-2)(4x+5)=0

=>3x-2=0 hoặc 4x+5=0

=>x=2/3 hoặc x=-5/4

b: (2,3x-6,9)(0,1x+2)=0

=>2,3x-6,9=0 hoặc 0,1x+2=0

=>x=3 hoặc x=-20

c: =>(x-3)(2x+5)=0

=>x-3=0 hoặc 2x+5=0

=>x=3 hoặc x=-5/2

$\textbf{a)}$

$(3x-2)(4x+5)=0$

$\Leftrightarrow3x-2=0$ hoặc $4x+5=0$

$\Leftrightarrow x=\dfrac23$ hoặc $x=-\dfrac54.$

=>15(2x-1)-2(3x+1)+20=8(3x+2)

=>30x-15-6x-2+20=24x+16

=>24x+3=24x+16

=>Loại

thank youu

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)

\(\Leftrightarrow x-1=3x-2\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

 \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)

\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)

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

c: =>(x-3)(x²+3x+5)=0

=>x-3=0

hay x=3

d: =>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x-3)(x-4)=0

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

a: \(\left|-5x\right|+\left|7-x\right|=27\)

=>5|x|+|x-7|=27(1)

TH1: x<0

=>x<0; x-7<0

(1) sẽ trở thành: -5x+7-x=27

=>-6x=20

=>\(x=\frac{20}{-6}=-\frac{10}{3}\) (nhận)

TH2: 0<=x<7

=>x>=0; x-7<0

(1) sẽ trở thành: 5x+7-x=27

=>4x=20

=>x=5(nhận)

TH3: x>=7

=>x>0; x-7>=0

(1) sẽ trở thành: 5x+x-7=27

=>6x=34

=>x=34/6=17/3(loại)

b: |3x-1|+|x+4|=21(2)

TH1: x<-4

=>x+4<0; 3x-1<0

(2) sẽ trở thành: 1-3x-x-4=21

=>-4x-3=21

=>-4x=24

=>x=-6(nhận)

TH2: -4<=x<1/3

=>x+4>=0; 3x-1<0

(2) sẽ trở thành: 1-3x+x+4=21

=>-2x+5=21

=>-2x=16

=>x=-8(loại)

TH3: x>=1/3

=>x+4>0; 3x-1>=0

(2) sẽ trở thành: 3x-1+x+4=21

=>4x+3=21

=>4x=18

=>x=18/4=9/2(nhận)

c: \(2\left|3x\right|-5\left|x+2\right|=-7\)

=>6|x|-5|x+2|=-7

=>\(5\left|x+2\right|-6\left|x\right|=7\) (3)

TH1: x<-2

=>x+2<0; x<0

(3) sẽ trở thành: 5(-x-2)-6(-x)=7

=>-5x-10+6x=7

=>x=17(loại)

TH2: -2<=x<0

=>x+2>=0; x<0

(3) sẽ trở thành: 5(x+2)-6(-x)=7

=>5x+10+6x=7

=>11x=-3

=>x=-3/11(nhận)

TH3: x>=0

=>x+2>0; x>=0

(3) sẽ trở thành: 5(x+2)-6x=7

=>5x+10-6x=7

=>10-x=7

=>x=3(nhận)