Giải thích các bước giải:
b.$\dfrac{x-4}{96}+\dfrac{x-3}{97}=\dfrac{x-2}{98}+\dfrac{x-1}{99}$
$\to (\dfrac{x-4}{96}-1)+(\dfrac{x-3}{97}-1)=(\dfrac{x-2}{98}-1)+(\dfrac{x-1}{99}-1)$
$\to\dfrac{x-100}{96}+\dfrac{x-100}{97}=\dfrac{x-100}{98}+\dfrac{x-100}{99}$
$\to x-100=0\to x=100$
c.$(x^2-5x+6)^3+(2-x^2)^2=(8-5x)^3$
$\to (x^2-5x+6)^3-(8-5x)^3+(x^2-2)^2=0$
$\to (x^2-5x+6-8+5x)((x^2-5x+6)^2+(x^2-5x+6)(8-5x)+(8-5x)^2)+(x^2-2)^2=0$
$\to (x^2-2)((x^2-5x+6)^2+(x^2-5x+6)(8-5x)+(8-5x)^2)+(x^2-2)^2=0$
$\to (x^2-2)((x^2-5x+6)^2+(x^2-5x+6)(8-5x)+(8-5x)^2+x^2-2)=0$
$\to x^2-2=0$
$\to x=\pm\sqrt{2}$
