Đáp án: a.$x=\dfrac{-13\pm\sqrt{41}}{2}$
b.$x=\dfrac{-119\sqrt{12865}}{12}$
Giải thích các bước giải:
a.Ta có:
$(x^2+2x+4)(2-x)+x(x-3)(x+4)-x+24=0$
$\to (2-x)(2^2+2x+x^2)+x(x^2+x-12)-x+24=0$
$\to 2^3-x^3+x^3+x^2-12x-x+24=0$
$\to x^2-13x+32=0$
$\to x^2-2\cdot x\cdot \dfrac{13}{2}+( \dfrac{13}{2})^2+32-( \dfrac{13}{2})^2=0$
$\to (x+\dfrac{13}{2})^2-\dfrac{41}{4}=0$
$\to (x+\dfrac{13}{2})^2=\dfrac{41}{4}$
$\to x+\dfrac{13}{2}=\dfrac{\pm\sqrt{41}}{2}$
$\to x=\dfrac{-13\pm\sqrt{41}}{2}$
b.Ta có:
$(\dfrac{x}{2}+3)(5-6x)+(12x-2)(\dfrac{x}{3}+3)=0$
$\to \dfrac52x-3x^2+15-18x+4x^2+36x-\dfrac23x-6=0$
$\to x^2+\dfrac{119}{6}x+9=0$
$\to x^2+2\cdot x\cdot\dfrac{119}{12}+(\dfrac{119}{12})^2+9-(\dfrac{119}{12})^2=0$
$\to (x+\dfrac{119}{12})^2-\dfrac{12865}{144}=0$
$\to (x+\dfrac{119}{12})^2=\dfrac{12865}{144}$
$\to x+\dfrac{119}{12}=\dfrac{\sqrt{12865}}{12}$
$\to x=\dfrac{-119\sqrt{12865}}{12}$
