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

 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

Điều kiện: $x\ne-2,\,-5,\,-10,\,-17,\,-12.$

$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\left(\dfrac1{x+2}-\dfrac1{x+5}\right)+\left(\dfrac1{x+5}-\dfrac1{x+10}\right)+\left(\dfrac1{x+10}-\dfrac1{x+17}\right)=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow15(x+12)=x(x+2)$

$\Leftrightarrow x^2-13x-180=0$

$\Leftrightarrow(x-20)(x+9)=0$

$\Leftrightarrow x=20\ \text{hoặc}\ x=-9.$

Vì $20,\,-9$ đều thỏa mãn điều kiện xác định nên $x=20\ \text{hoặc}\ x=-9.$

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

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

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

<=> x = 15