Question

Answer 1

`8/(3x-1)=3x+1`                  ĐKXĐ : `x \ne 1/3`

`<=>8=(3x+1)(3x-1)`

`<=>3x^2-1=8`

`<=>3x^2=9`

`<=>x^2=3`

`<=>x=\pm \sqrt{3}` ( tm ĐKXĐ )

Vậy `S={\pm \sqrt{3}}`

Answer 2

Câu 1:

ĐKXĐ: $x\neq \frac{1}{3}$

PT $\Leftrightarrow 8=(3x+1)(3x-1)$

$\Leftrightarrow 8=9x^2-1$

$\Leftrightarrow 9=9x^2$

$\Leftrightarrow 1=x^2$

$\Leftrightarrow x=\pm 1$

Đáp án D.

Answer 3

Câu 2:

\(\frac{2-x}{3}< \frac{3-2x}{5}\Leftrightarrow 5(2-x)< 3(3-2x)\)

\(\Leftrightarrow 10-5x< 9-6x\Leftrightarrow x< -1\)

Đáp án C.

Answer 4

Câu 3:

\(A=\frac{x^2-x+3}{x-3}=\frac{x(x-3)+2(x-3)+9}{x-3}\)

\(=x+2+\frac{9}{x-3}=(x-3)+\frac{9}{x-3}+5\geq 2\sqrt{(x-3).\frac{9}{x-3}}+5\) (theo BĐT Cô-si)

\(=6+5=11\)

Đáp án C.

Answer 5

`8/(3x-1)=3x+1`                 ĐKXĐ : `x \ne 1/3`

`<=>8=(3x+1)(3x-1)`

`<=>8=9x^2-1`

`<=>9x^2-9=0`

`<=>x^2-1=0`

`<=>(x-1)(x+1)=0`

`<=>` \( \left[\begin{array}{} x-1=0\\ x+1=0 \end{array} \right. \)

`<=>` \( \left[\begin{array}{} x=1 \ \rm(tm)\\ x=-1 \ \rm(tm) \end{array} \right.\)

Vậy `S={\pm 1}`

Answer 6

`(2-x)/3<(3-2x)/5`

`<=>(5(2-x))/15<(3(3-2x))/15`

`<=>5(2-x)<3(3-2x)`

`<=>10-5x<9-6x`

`<=>-5x+6x<9-10`

`<=>x<-1`

`->` Chọn C

Answer 7

Câu 1: D

Câu 2: C

Câu 3: C