Giải thích các bước giải:
a.Ta có :
$\dfrac{(x-2)^2}{12}-\dfrac{(x+1)^2}{21}=\dfrac{(x-9)(x-6)}{28}$
$\to 7\left(x-2\right)^2-4\left(x+1\right)^2=3\left(x-9\right)\left(x-6\right)$
$\to 3x^2-36x+24=3\left(x^2-15x+54\right)$
$\to 3x^2-36x+24=3x^2-45x+162$
$\to 9x=138$
$\to x=\dfrac{46}{3}$
b.Ta có :
$\dfrac{3(2x+1)}{4}-\dfrac{5x+3}{6+x}+\dfrac{1}{3}=\dfrac{x+7}{12}$
$\to 9\left(2x+1\right)\left(x+6\right)-12\left(5x+3\right)+4\left(x+6\right)=\left(x+7\right)\left(x+6\right)$
$\to 18x^2+61x+42=x^2+13x+42$
$\to 17x^2+48x=0$
$\to x(17x+48)=0$
$\to x=0$ hoặc $17x+48=0\to x=-\dfrac{48}{17}$
c.Ta có :
$2x+4(x-2)=5$
$\to 2x+4x-8=5$
$\to 6x=13$
$\to x=\dfrac{13}{6}$
