Question

Câu 15. (1 điểm) Tìm $x$, biết:

a) $3x\left( x-1 \right)-1+x=0$;

b) ${{x}^{2}}-9x=0$.

Answer 1

 

Answer 2

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

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

<=>x=1 ;x= \(\dfrac{1}{3}\)

b)x²-9x=0<=>x.(x-9)=0

=>x=0 ; x-9=0

<=>x=0 ; x=9

Answer 3

a) 3x (x - 1) - 1 + x = 0
    3x (x - 1) - (x - 1) = 0
    (x - 1)(3x - 1) = 0
=> x - 1 = 0 hoặc 3x - 1 = 0
=> x = 1      hoặc x = 1/3

b) x² - 9x = 0
    x(x - 9) = 0
=> x = 0 hoặc x - 9 = 0
=> x =0 hoặc x = 9

Answer 4

a) x = 1/3 ; x= 1                    b) x=0 ; x=9

Answer 5

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 6

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 7

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 8

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 9

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 10

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 11

a) 3�(�−1)−1+�=0$3x\left(x-1\right)-1+x=0$

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

       ⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)

       ⇒TH2:x-1=0⇒x=1

Vậy xϵ{-\(\dfrac{1}{3}\);1}

b) �2−9�=0x²
−9x=0

x(x-9)=0

         ⇒TH1:x=0

         ⇒TH2:x-9=0⇒x=9

Vậy xϵ{0;9}

Answer 12

 

Answer 13

Answer 14

Answer 15

Answer 16

 

Answer 17

 

Answer 18

 

Answer 19

 

Answer 20

 

Answer 21

  

Answer 22

 

Answer 23

 

Answer 24

 

Answer 25

a) $3x\left(x-1\right)-1+x=0$

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

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

Suy ra $3x+1=0$ hoặc $x-1=0$

Vậy x =1/3 hoặc x =1

b)x^2$-9x=0$

$x\left(x-9\right)=0$

Suy ra $x=0$ hoặc $x=9$.

Answer 26

a) $3x\left(x-1\right)-1+x=0$

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

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

Suy ra $3x+1=0$ hoặc $x-1=0$

Vậy �=−13x=−31​ hoặc $x=1$

b) �2−9�=0x2−9x=0

$x\left(x-9\right)=0$

Suy ra $x=0$ hoặc $x=9$.

Answer 27

Answer 28

a) $3x\left(x-1\right)-1+x=0$

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

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

Suy ra $3x+1=0$ hoặc $x-1=0$

Vậy �=−13x=−31​ hoặc $x=1$

b) �2−9�=0x2−9x=0

$x\left(x-9\right)=0$

Suy ra $x=0$ hoặc $x=9$.

Answer 29

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

(3x+1)(x−1)=0

TH1:3x+1=0

x=-1/3

TH2:x-1=0

x=1

Vậy x thuộc {-1/3;1}

b)x^2-9x=0

x(x-9)=0

TH1:x=0

TH2:x-9=0

x=9

Vậy x thuộc {0;9}

 

Answer 30