a) 2x²-4x-x+2=0

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

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

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

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

b) 3x²-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

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

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

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

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

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

\(\Leftrightarrow5\left(x-4\right)-\left(x-4\right)^2=0\\ \Leftrightarrow\left(x-4\right)\left(5-x+4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(9-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=9\end{matrix}\right.\)

\(-x^4+4x^2-5x^2+20=0\\\Rightarrow -(x^4-4x^2)-(5x^2-20)=0\\\Rightarrow-x^2(x^2-4)-5(x^2-4)=0\\\Rightarrow(x^2-4)(-x^2-5)=0\\\Rightarrow-(x-2)(x+2)(x^2+5)=0\\\Rightarrow(2-x)(x+2)=0(vì.x^2+5>0\forall x)\)

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

x=-5,16769911504425

a) \(5x - 30 = 0\)

\(5x = 0 + 30\)     

\(5x = 30\)

\(x = 30:5\)

\(x = 6\)      

Vậy phương trình có nghiệm \(x = 6\).

b) \(4 - 3x = 11\)

\( - 3x = 11 - 4\)

\( - 3x =  7\)

\(x = \left( { 7} \right):\left( { - 3} \right)\)

\(x = \dfrac{-7}{3}\)

Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).

c) \(3x + x + 20 = 0\)               

\(4x + 20 = 0\)

\(4x = 0 - 20\)

\(4x =  - 20\)

\(x = \left( { - 20} \right):4\)

\(x =  - 5\)   

Vậy phương trình có nghiệm \(x =  - 5\).

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

\(\dfrac{1}{3}x - x = 2 - \dfrac{1}{2}\)

\(\dfrac{{ - 2}}{3}x = \dfrac{3}{2}\)

\(x = \dfrac{3}{2}:\left( {\dfrac{{ - 2}}{3}} \right)\)

\(x = \dfrac{{ - 9}}{4}\)

Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).

a,  3 ( x + 1 ) - 2 ( 3 x - 4 ) = - 13

=> 3x + 3 - 6x + 8 = - 13

=> 6x - 3x = 3 + 8 + 13

=> 3x = 24

=> x = 8

b, 2 ( x - 3 ) - 4 ( 2 x - 1 ) = - 20

=> 2x - 6 - 8x + 4 = - 20

=> 8x - 2x = - 6 + 4 + 20

=> 6x = 18

=> x = 3

c, 2 x ( x + 3 ) = 0

=> \(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

d, ( x - 1 ) ( 5 x - x ) = 0

=> \(\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

e, ( x + 3 ) ² ( 4 - x ) = 0

=> \(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x+3=0\\4-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)

a) \(3\left(x+1\right)-2\left(3x-4\right)=-13\)

\(\Leftrightarrow3x+3-6x+8=-13\)

\(\Leftrightarrow3x-6x=-13-3-8\)

\(\Leftrightarrow-3x=-24\)

\(\Leftrightarrow x=8\)

Vậy \(x=8\)

b) \(2\left(x-3\right)-4\left(2x-1\right)=-20\)

\(\Leftrightarrow2x-6-8x+4=-20\)

\(\Leftrightarrow2x-8x=-20+6-4\)

\(\Leftrightarrow-6x=-18\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

c) \(2x\left(x+3\right)=0\)

\(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

d)\(\left(x-1\right)\left(5x-x\right)=0\)

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

Vậy \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

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

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

Vậy \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)

cảm ơn bạn nhé hi hi

Bài 4 :

\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)

\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)

Bài 5 :

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

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

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

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

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

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

Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)

Bài 6 :

\(\left(4x+20\right)-\left(x+5\right)^2=0\)

\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

Vậy \(x=-5\) hoặc \(x=-1\)

\(1,5x^2-4\left(x^2-2x+1\right)+20=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4+20=0\)

\(\Leftrightarrow x^2+8x+16=0\)

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

\(\Rightarrow x+4=0\Rightarrow x=-4\)

\(2,x\left(x-2\right)-5x+10=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)