Giải thích các bước giải:
$a)\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}$
$⇒\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0$
$⇒(x-23)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0$
Ta thấy $\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\neq0$
$⇒x-23=0$
$⇒x=23$
Vậy $x=23$
$b)\left|\dfrac{x+2}{98}+1\right|+\left|\dfrac{x+3}{97}+1\right|=\left|\dfrac{x+4}{96}+1\right|+\left|\dfrac{x+5}{95}+1\right|$
$⇒\left|\dfrac{x+100}{98}\right|+\left|\dfrac{x+100}{97}\right|=\left|\dfrac{x+100}{96}\right|+\left|\dfrac{x+100}{95}\right|$
$⇒\left|\dfrac{x+100}{98}\right|+\left|\dfrac{x+100}{97}\right|-\left|\dfrac{x+100}{96}\right|-\left|\dfrac{x+100}{95}\right|=0$
$⇒|x+100|\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0$
Ta thấy $\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\neq0$
$⇒|x+100|=0$
$⇒x=-100$
Vậy $x=-100$
$c)\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}$
$⇒\left(\dfrac{x+1}{1998}+1\right)+\left(\dfrac{x+2}{1997}+1\right)=\left(\dfrac{x+3}{1996}+1\right)+\left(\dfrac{x+4}{1995}+1\right)$
$⇒\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}$
$⇒\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0$
$⇒(x+1999)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0$
Ta thấy $\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\neq0$
$⇒x+1999=0$
$⇒x=-1999$
Vậy $x=-1999$
Giải thích:
Một tích bằng 0 khi một trong hai thừa số bằng 0.
Ví dụ:
$(a+b)(c+d)=0$
$⇒\left[ \begin{array}{l}a+b=0\\c+d=0\end{array} \right.$
