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}

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}

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}

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}

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}

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}

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}

a)x²+25-10x

=(x+5)²

b)-8y³+x³

=(-2y+x)(4y²+2xy+x³)