1) x+5 / 3x-6 - 1/2 = 2x-3 / 2x-4 = 0 2) 12 / 1-9x^2 = 1-3x / 1+3x - 1+3x / 1-3x (đề bài là giải phương trình ạ ) ( dấu / là phần , dấu ^ là mũ ạ )

Đáp án:

$1)$ S = `{16}`

$2)$ S = `{-1}`

Giải thích các bước giải:

$1)$ $\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4\:}=0$

$⇔-x+16=0$

$⇔16−x=0$

$⇔-x=-16$

$⇔x=16$

Vậy S = `{16}`

$2)$ $\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$ `;(x!=+-1/3)`

$⇔\dfrac{12}{\left(1+3x\right)\left(1-3x\right)\:}=\dfrac{\:1-3x}{1+3x\:}-\dfrac{1+3x}{\:1-3x\:\:\:}$

$⇔12=\left(1-3x\right)^2-\left(1+3x\right)^2$

$⇔\left(1-3x\right)^2-\left(1+3x\right)^2=12$

$⇔-12x=12$

$⇔x=-1$

Vậy S = `{-1}`

Đáp án:

$\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4\:}=0$

$⇔-x+16=0$

$⇔16−x=0$

$⇔x=16$

Vậy `S =` `{16}`

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$\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$ `;(x!=+-1/3)`

$⇔\dfrac{12}{\left(1+3x\right)\left(1-3x\right)\:}=\dfrac{\:1-3x}{1+3x\:}-\dfrac{1+3x}{\:1-3x\:\:\:}$

$⇔12=\left(1-3x\right)^2-\left(1+3x\right)^2$

$⇔\left(1-3x\right)^2-\left(1+3x\right)^2=12$

$⇔-12x=12$

$⇔x=12:(-12)$

$⇔x=-1$

Vậy `S =` `{-1}`