c, -x + $\dfrac{4}{5}$ = $\dfrac{1}{2}$
⇔ -x = $\dfrac{1}{2}$ - $\dfrac{4}{5}$
⇔ -x = $\dfrac{5}{10}$ - $\dfrac{8}{10}$
⇔ -x = $\dfrac{-3}{10}$
⇔ x = $\dfrac{3}{10}$
Vậy $x$ = $\dfrac{3}{10}$
d, $\dfrac{1}{x. ( x+ 1)}$ + $\dfrac{1}{( x+ 1). ( x+ 2)}$ + $\dfrac{1}{( x+ 2). ( x+ 3)}$ $-$ $\dfrac{1}{x}$ $=$ $\dfrac{1}{2010}$
⇔ $\dfrac{1}{x}$ $-$ $\dfrac{1}{x+ 1}$ + $\dfrac{1}{x+ 1}$ $-$ $\dfrac{1}{x +2}$ +$\dfrac{1}{x+ 2}$ $-$ $\dfrac{1}{x+3}$ $-$ $\dfrac{1}{x}$ = $\dfrac{1}{2010}$
⇔ $\dfrac{1}{x}$ $-$ $\dfrac{1}{x+ 3}$ $-$ $\dfrac{1}{x }$ = $\dfrac{1}{2010}$
⇔ ( $\dfrac{1}{x}$ $-$ $\dfrac{1}{x }$) $-$ $\dfrac{1}{x+ 3}$ = $\dfrac{1}{2010}$
⇔ $0$ - $\dfrac{1}{x+ 3}$ $=$ $\dfrac{1}{2010}$
⇔ $\dfrac{-1}{x+ 3}$ = $\dfrac{-1}{-2010}$
⇔ x+ 3= -2010
⇔ x= -2013
Vậy x= -2013
A = $1$ $-$ 2/ 3.5 - 2/5.7 -...- 2/63. 65
A= $1$ - ( 2/3.5+ 2/5.7+...+ 2/63. 65)
A= $1$ - $\dfrac{1}{3}$ $-$ $\dfrac{1}{5}$ $+$ $\dfrac{1}{5}$ $-$ $\dfrac{1}{7}$ $+$...$+$ $\dfrac{1}{63}$ $-$ $\dfrac{1}{65}$
$A$ $=$ $1$ $-$ ($\dfrac{1}{3}$ $-$ $\dfrac{1}{65}$ )
$A$ $=$ $1$ $-$ $\dfrac{62}{195}$
$A$ $=$ $\dfrac{133}{195}$
Vậy $A$ $=$ $\dfrac{133}{195}$
Chúc học tốt! Nhớ kiểm tra lại nhé!
