Answer:

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

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

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

\(x^2-5x+6=0\)

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

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

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

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

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

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

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

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

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

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

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)

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

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

d) Ta có: \(5x^2+x=0\)

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

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

b) \(x\left(2x+5\right)=0\)

TH1: \(x=0\)

TH2: \(2x+5=0\Leftrightarrow2x=-5\Leftrightarrow x=\dfrac{-5}{2}\)

\(a,\Rightarrow5x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x-3\right)\left(5x-x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(4x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{4}\end{matrix}\right.\\ b,\Rightarrow x\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

5x(x – 3) – x + 3 = 0

⇔ 5x(x – 3) – (x – 3) = 0

(Xuất hiện nhân tử chung x – 3)

⇔ (x – 3)(5x – 1) = 0

⇔ x – 3 = 0 hoặc 5x – 1= 0

+ x – 3 = 0 ⇔ x = 3

+ 5x – 1 = 0 ⇔ 5x = 1 ⇔ x = 1/5

Vậy x = 3 hoặc x = 1/5.

Mình trình bày trong hình ^^ Bn tham khảo nhé

d: Ta có: \(9x^2+6x-8=0\)

\(\Leftrightarrow9x^2+12x-6x-8=0\)

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

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

e: Ta có: \(x\left(x-2\right)+x-2=0\)

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

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

f: Ta có: \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

a) 5x( x - 1 ) = x - 1

<=> 5x² - 5x = x - 1

<=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0

<=> 5x² - 5x - x + 1 = 0

<=> 5x( x - 1 ) - 1( x - 1 ) = 0

<=> ( x - 1 )( 5x - 1 ) = 0

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

b) 2( x + 5 ) - x² - 5x = 0

<=> 2x + 10 - x² - 5x = 0

<=> -x² - 3x + 10 = 0

<=> -x² - 5x + 2x + 10 = 0

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

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

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

c) x² - 2x - 3 = 0

<=> x² + x - 3x - 3 = 0

<=> x( x + 1 ) - 3( x + 1 ) = 0

<=> ( x + 1 )( x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

d) 2x² + 5x - 3 = 0

<=> 2x² - x + 6x - 3 = 0

,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0

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

<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

a) 5x ( x - 1 ) = x - 1 <=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0 <=> 5x² - x - ( 5x - 1 ) = 0 

<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0

<=> x = 1 hoặc x = 1/5

b) 2 ( x + 5 ) - x² - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0

<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5

c) x² - 2x - 3 = 0 <=> x² + x - 3x - 3 = 0 

<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0 

<=> x = 3 hoặc x = -1

d) 2x²  + 5x - 3 = 0

Ta có : delta = 5² - 4.2.3 = 25 - 24 = 1

Khi đó : x = -1 hoặc x = 3/2  

a) 5x(x-1)=x-1

\(\Leftrightarrow\)5x(x-1)-(x-1)=0

\(\Leftrightarrow\)(x-1)(5x-1)=0

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

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

vậy x=1 hoặc x=1/5

b) 2(x+5)-x²-5x=0

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

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

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

Vậy x=2 hoặc x=-5

c) x²-2x-3=0

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

vậy...

\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)

12(x-1)=0

     (x-1)=0:12

      x-1=0

      x=0+1

      x= 1

Vậy x= 1

12 ( x - 1 ) = 0
       x - 1   = 0 : 12
       x - 1   = 0
=>   x        = 1

Ko viết lại đề

Câu 1: chia ra làm 3 trường hợp

Câu 2: 

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

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

\(\Rightarrow x+2=0\)

\(x=-2\)

câu 1:suy ra 

5x=0vậy x=0

x-3=0vậy x=3

-2x+6=0vậy x=3