a) Bạn xem lại đề
b) $(2x-4)(4+7x)=0$
$2(x-2)(4+7x)=0$
$\to$
Nếu:
$(x-2)=0 \to x=2$
$4+7x=0\to x=\dfrac{-4}{7}$
c) $\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}$
$\to \dfrac{x-1}{x}+\dfrac{1}{x+1}-\dfrac{2x-1}{x(x+1)}=0$
$\to \dfrac{(x-1)(x+1)+x-2x+1}{x(x+1)}=0$
$\to \dfrac{x^2-1-x+1}{x(x+1)}=0$
$\to \dfrac{x^2-x}{x(x+1)}=0$
$\to x^2 -x=0$
$\to x(x-1)=0$
$\to x=0; x=1$
$\to S={0;1}$
d) $\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}$
$\to (\dfrac{x+1}{65}+1)+(\dfrac{x+3}{63}+1)=(\dfrac{x+5}{61}+1)+(\dfrac{x+7}{59}+1)$
$\to \dfrac{x+1+65}{65}+\dfrac{x+3+63}{63}-\dfrac{x+5+61}{61}-\dfrac{x+7+59}{59}=0$
$\to \dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0$
$\to (x+66)(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59})=0$
Do:
$\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0$
$\to x+66=0$
$\to x=-66$
