Giải phương trình sau: $\frac{(x+10)(x+4)}{12}$$-$$\frac{(x+4)(2-x)}{4}$$=$$\frac{(x+10)(x-2)}{3}$ 09/11/2021 Bởi Maria Giải phương trình sau: $\frac{(x+10)(x+4)}{12}$$-$$\frac{(x+4)(2-x)}{4}$$=$$\frac{(x+10)(x-2)}{3}$
$\dfrac{(x+10)(x+4)}{12}-\dfrac{(x+4)(2-x)}{4}=\dfrac{(x+10)(x-2)}{3}$ $\to \dfrac{(x+10)(x+4)-3(x+4)(2-x)}{12}-\dfrac{4(x+10)(x-2)}{12}=0$ $\to (x+10)(x+4)- 3(x+4)(2-x)-4(x+10)(x-2)=0$ $\to x^2+14x+40-3(-x^2-2x+8)-4(x^2+8x-20)=0$ $\to x^2+14x+40+3x^2+6x-24-4(x^2+8x-20)=0$ $\to 4x^2 +20x+16-4x^2-32x+80=0$ $\to 96-12x=0$ $\to x=8$ Bình luận
`((x+10)(x+4))/12-((x+4)(2-x))/4=((x+10)(x-2))/3` `<=>((x+10)(x+4))/12+(3(x+4)(x-2))/12=(4(x+10)(x-2))/12` `<=>((x+10)(x+4)+3(x+4)(x-2))/12=(4(x+10)(x-2))/12` `<=>(x+10)(x+4)+3(x+4)(x-2)=4(x+10)(x-2)` `<=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80` `<=>x^2+14x+40+3x^2+6x-24-4x^2-32x+80=0` `<=>-12x+96=0` `<=>-12x=-96` `<=>x=8` Vậy phương trình có nghiệm duy nhất`S={8}` Bình luận
$\dfrac{(x+10)(x+4)}{12}-\dfrac{(x+4)(2-x)}{4}=\dfrac{(x+10)(x-2)}{3}$
$\to \dfrac{(x+10)(x+4)-3(x+4)(2-x)}{12}-\dfrac{4(x+10)(x-2)}{12}=0$
$\to (x+10)(x+4)- 3(x+4)(2-x)-4(x+10)(x-2)=0$
$\to x^2+14x+40-3(-x^2-2x+8)-4(x^2+8x-20)=0$
$\to x^2+14x+40+3x^2+6x-24-4(x^2+8x-20)=0$
$\to 4x^2 +20x+16-4x^2-32x+80=0$
$\to 96-12x=0$
$\to x=8$
`((x+10)(x+4))/12-((x+4)(2-x))/4=((x+10)(x-2))/3`
`<=>((x+10)(x+4))/12+(3(x+4)(x-2))/12=(4(x+10)(x-2))/12`
`<=>((x+10)(x+4)+3(x+4)(x-2))/12=(4(x+10)(x-2))/12`
`<=>(x+10)(x+4)+3(x+4)(x-2)=4(x+10)(x-2)`
`<=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80`
`<=>x^2+14x+40+3x^2+6x-24-4x^2-32x+80=0`
`<=>-12x+96=0`
`<=>-12x=-96`
`<=>x=8`
Vậy phương trình có nghiệm duy nhất`S={8}`