a: f(x)=0

=>x(2x-1)=0

=>x=0 hoặc x=1/2

b: g(x)=0

=>x^2-1=0

=>x^2=1

=>x=1 hoặc x=-1

c: h(x)=0

=>x^2-3=0

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

 720 : ( x . 2 + x . 3 ) = 3.2
720 : ( x . 2 + x.3 ) = 6
( x .2 + x.3 )           = 720 : 6 
x.2+x.3 = 120
x . ( 2 + 3 ) = 120
x . 5 = 120
     x     = 120 : 5 
    x      = 24

x^2=16

Suy ra x^2=4^2

Vậy x^2=4^2

[1/(x+2)-1/(x+5)]+[1/(x+5)-1/(x+10)]+[1/(x+10)-1/(x+17)]=x/15.[1/(x+2)-1/(x+17)]
1/(x+2)-1/(x+17)=x/15.[1/(x+2)-1/(x+17)]
1=x/15
x=15

a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)

=>1/x-1=3/4

=>x-1=4/3

=>x=7/3

a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)

=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)

=>\(6x-\dfrac{39}{4}=1\)

=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)

=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)

b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)

=>\(2x-6=3x+6-x+1\)

=>2x-6=2x+7

=>-6=7(vô lý)

c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)

=>\(x^2+3x+x^2-2x=2x^2-2x\)

=>3x-2x=-2x

=>3x=0

=>x=0

d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)

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

=>\(3x^2-7x-4-x^2+x=0\)

=>\(2x^2-6x-4=0\)

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

=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)