1) (2x-1)(x+3)(2-x)=0

=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0

=>x=1/2 hoặc x=-3 hoặc x=2

2)x^3 + x^2 + x + 1 = 0

=>.x^2(x+1)+(x+1)=0

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

=>x^2+1=0 hoặc x+1=0 

=>                      x =-1

3) 2x(x-3)+5(x-3) =0    

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

=>2x+5=0 hoặc x-3=0

=>x=-5/2 hoặc x=3

4)x(2x-7)-(4x-14)=0

=> (x-2)(2x-7)=0

=> x-2 =0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

5)2x^3+3x^2+2x+3=0

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

=>(x^2+1)(2x+3)=0

=>x^2+1=0 hoặc 2x+3=0

=>                      x =-3/2

x = 3/2 đó mình chắc chắn 100 %

X= 3/2 nha bạn, chắc chắn đó.

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

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

\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)

\(\Leftrightarrow-x^3+5x^2-5x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)

a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)

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

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

\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)

=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)

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

\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)

\(\Leftrightarrow-12=0\left(vn\right)\)

c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)

\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)

\(\Leftrightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

a) x( x - 2 ) - x( x - 1 )( x - 3 ) = 0

<=> x² - 2x - x( x² - 4x + 3 ) = 0

<=> x² - 2x - x³ + 4x² - 3x = 0

<=> -x³ + 5x² - 5x = 0

<=> -x( x² - 5x + 5 ) = 0

<=> \(\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\)

+) -x = 0 => x = 0

+) x² - 5x + 5 = 0 (*)

\(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot1\cdot5==25-20=5\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{5}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{5}}{2}\end{cases}}\)

Vậy S = { \(0;\frac{5\pm\sqrt{5}}{2}\)}

b) ( 2x - 5 )( x + 3 ) - ( x - 1 )( 2x + 3 ) = 0

<=> 2x² + x - 15 - ( 2x² + x - 3 ) = 0

<=> 2x² + x - 15 - 2x² - x + 3 = 0

<=> -12 = 0 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x - 2 )( x² + 2x + 8 ) - x³ - 2x + 1 = 0

<=> x³ + 4x - 16 - x³ - 2x + 1 = 0

<=> 2x - 15 = 0

<=> 2x = 15

<=> x = 15/2

\(a.2x-3=4x+6\) 

\(\Leftrightarrow2x-3-4x-6=0\) 

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow x=\dfrac{9}{2}\)

\(S=\left\{\dfrac{9}{2}\right\}\) 

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

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

\(\Leftrightarrow2x^2+2=0\)

\(\Leftrightarrow x\left(2x+2\right)=0\) 

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

\(S=\left\{0,-1\right\}\) 

Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm

a: =>-2x=9

=>x=-9/2

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=0 hoặc x=-1

\(a,2x-3=4x+6\)

\(\Leftrightarrow2x-4x=6+3\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

\(b,\) Ghi vậy mình không làm được.

\(c,\)\(x\left(x-1\right)+x\left(x+3\right)=0\)

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

\(\Leftrightarrow x\left(2x+2\right)=0\)

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

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

\(d,\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}=0\left(dkxd:x\ne-1;x\ne3\right)\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x-3\right)-2.2}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow x^2+x-x^2+3x-4=0\)
\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow x=1\left(tmdk\right)\)

Vậy \(S=\left\{1\right\}\)

`a,(2x-5)(12+5x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

`b, (x-3)(x-4)-2(x-3)=0`

`<=>(x-3)(x-4-2)=0`

`<=>(x-3)(x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

`c, x(x-1)(x+1)=0`

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

`d, (2x)/3 +(2x-1)/6=0`

`<=> (4x)/6 +(2x-1)/6=0`

`<=> (4x+2x-1)/6=0`

`<=> (6x-1)/6=0`

`<=> 6x-1=0`

`<=> 6x=1`

`<=>x=1/6` ( đề là vậy à bạn )

 a) \(\left(2x-5\right)\left(12+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

c) \(x\left(x-1\right)\left(x+1\right)=0\)

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

d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)

\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)

\(\Leftrightarrow6x-1=0\)

\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)

a) 3x(4x - 3) - 2x(5 - 6x) = 0

=> 6x² - 9x - 10x + 12x² = 0

=> 18x² - 19x = 0

=> x(18x - 19) = 0

=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)

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

=> 10x - 15 + 4x² - 8x + 6x - 4x² = 0

=> 8x - 15 = 0

=> 8x = 15

=> x = 15 : 8 = 15/8

c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)

=> 6x - 3x² + 2x² - 2x = 5x² + 15x

=> 4x - x² - 5x² - 15x = 0

=> -6x² - 11x = 0

=> -x(6x - 11) = 0

=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)

a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)

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

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)

d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

Mà \(14x^2+18>0\)nên pt vô nghiệm

a) Ta có: 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)

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

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

hay \(x=\frac{15}{8}\)

Vậy: \(x=\frac{15}{8}\)

c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)

\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)

\(\Leftrightarrow-6x^2-11x=0\)

\(\Leftrightarrow6x^2+11x=0\)

\(\Leftrightarrow x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)

d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

\(\Leftrightarrow14x^2=-18\)

mà \(14x^2\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

a/

\(\Leftrightarrow x-2x^2+2x^2-3x-4x+6=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow x=1\)

b/

\(\Leftrightarrow2x^2-4x-2x^2-6x=0\)

\(\Leftrightarrow-10x=0\)

\(\Leftrightarrow x=0\)

c/

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

\(\Leftrightarrow3x\left(2x+3\right)=0\)

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

c/

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(9y^2+30y+25\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(3y+5\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\3x+5=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=-\frac{5}{3}\)

d/

\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1-2\left(4x^2-2x-2\right)+x=12\)

\(\Leftrightarrow8x^2+x+2-8x^2+4x+4=12\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)