\(1,\)

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

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

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

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

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

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

\(3,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

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

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

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

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

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

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

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

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

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

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

1C

2A

1C        2A

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

|x - 4| + |6 - x| = 0

|x  - 4| ; |6 - x| \(\ge\) 0

=> |x - 4| = |6 - x| = 0

|x - 4| = 0 => x= 4

|6 - x| = 0 => x=  6

Vì \(4\ne6\) n ê n không có giá trị của x

Bạn làm các câu khác tương tự 

a) x(x - 1) = 0

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

b) 3x² - 6x = 0

=> 3x.(x - 2) = 0

=> x.(x - 2) = 0

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

c) x(x - 6) + 10(x - 6) = 0

=> (x - 6)(x + 10) = 0

=> \(\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)

d) x³ - x = 0

=> x.(x² - 1) = 0

=> x.(x - 1).(x + 1) = 0

=> \(\left[\begin{array}{nghiempt}x=0\\x-1=0\\x+1=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=0\\x=1\\x=-1\end{array}\right.\)

a) 

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

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

Vậy x=0 ; x =1

b)

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

\(\Rightarrow3x\left(x-2\right)=0\)

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

Vậy x=0 ; x =2

c)

\(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Rightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)

Vậy x=6 ; x = -10

d)

\(x^3-x=0\)

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\\x+1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\\x=-1\end{array}\right.\)

Vậy x = 0 ; x = 1 ; x= - 1

a,\(x\left(x-1\right)=0\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)

b,\(3x^2-6x=0\)

\(\Rightarrow3x\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-2=0\end{array}\right.\)

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

c,\(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Rightarrow\left(x+10\right)\left(x-6\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-6=0\\x+10=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=6\\x=-10\end{array}\right.\)

d,\(x^3-x=0\)

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)

a)  x ( x + 6 ) = 0 ⇔ x = 0 x + 6 = 0 ⇔ x = 0 x = − 6

Vậy  x = 0 hoặc  x = - 6

b)  ( x − 3 ) . ( y + 7 ) = 0 ⇔ x − 3 = 0 y + 7 = 0 ⇔ x = 3 y = − 7

Vậy x = 3 hoặc x = -7

c)  ( x − 2 ) ( x 2 + 2 ) = 0 ⇔ x − 2 = 0 x 2 + 2 = 0 ⇔ x = 2 x 2 = − 2   ( L )

Vậy x = 2

\(a,\Leftrightarrow x^2-x-x^2+4x-4=2\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(6-x\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-6+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\\ c,\Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)