Giải phương trình tích a) __3x+2__ – __6__ = __9x^2__ 3x-2 2+3x 9x^2-4 12/11/2021 Bởi Valentina Giải phương trình tích a) __3x+2__ – __6__ = __9x^2__ 3x-2 2+3x 9x^2-4
$\frac{3x+2}{3x-2}$ – $\frac{6}{2+3x}$ = $\frac{9x^{2}}{9x^{2}-4}$ ⇔ $\frac{(3x+2)^{2}-6(3x-2)}{(3x-2)(3x+2)}$ = $\frac{9x^{2}}{(3x-2)(3x+2)}$ ⇔ $\frac{9x^{2}+12x+4-18x+12}{(3x-2)(3x+2)}$ = $\frac{9x^{2}}{(3x-2)(3x+2)}$ ⇔ $9x^{2}$+12x+4-18x+12 = $9x^{2}$ ⇔ 16 – 6x = 0 ⇔ -6x = -16 ⇔ x = $\frac{8}{3}$ Xin ctlhn ạ! Bình luận
Đáp án: `↓↓` Giải thích các bước giải: `(3x+2)/(3x-2) – 6/(2+3x)=(9x^2)/(9x^2-4)` $ĐKXĐ$`: x ne +-2/3` `=> (3x+2)/(3x-2) -6/(3x+2)=(9x^2)/((3x-2)(3x+2))` `=> ((3x+2)^2 – 6(3x-2))/((3x+2)(3x-2))=(9x^2)/((3x-2)(3x+2))` `=> (9x^2+12x+4-18x+12)/((3x+2)(3x-2))=(9x^2)/((3x+2)(3x-2))` `=> 9x^2-6x+16=9x^2` `=> -6x+16=0` `=> -6x=-16` `=> x=16/6=8/3` $(TMĐK)$ Bình luận
$\frac{3x+2}{3x-2}$ – $\frac{6}{2+3x}$ = $\frac{9x^{2}}{9x^{2}-4}$
⇔ $\frac{(3x+2)^{2}-6(3x-2)}{(3x-2)(3x+2)}$ = $\frac{9x^{2}}{(3x-2)(3x+2)}$
⇔ $\frac{9x^{2}+12x+4-18x+12}{(3x-2)(3x+2)}$ = $\frac{9x^{2}}{(3x-2)(3x+2)}$
⇔ $9x^{2}$+12x+4-18x+12 = $9x^{2}$
⇔ 16 – 6x = 0
⇔ -6x = -16
⇔ x = $\frac{8}{3}$
Xin ctlhn ạ!
Đáp án:
`↓↓`
Giải thích các bước giải:
`(3x+2)/(3x-2) – 6/(2+3x)=(9x^2)/(9x^2-4)`
$ĐKXĐ$`: x ne +-2/3`
`=> (3x+2)/(3x-2) -6/(3x+2)=(9x^2)/((3x-2)(3x+2))`
`=> ((3x+2)^2 – 6(3x-2))/((3x+2)(3x-2))=(9x^2)/((3x-2)(3x+2))`
`=> (9x^2+12x+4-18x+12)/((3x+2)(3x-2))=(9x^2)/((3x+2)(3x-2))`
`=> 9x^2-6x+16=9x^2`
`=> -6x+16=0`
`=> -6x=-16`
`=> x=16/6=8/3` $(TMĐK)$