Đáp án:

 `1\bb C`

`2\bb B`

Giải thích các bước giải:

 `\frac{x^4-8x^2-9}{x+3}=0`

ĐK: `x \ne -3`

`⇔ x^4-8x^2-9=0`

Đặt `x^2=t\ (t \ge 0)`

`⇔ t^2-8t-9=0`

`⇔ t^2+t-9t-9=0`

`⇔ (t+1)(t-9)=0`

`⇔` \(\left[ \begin{array}{l}t=-1\ (L)\\t=9\ (TM)\end{array} \right.\) 

+) `t=9 -> x^2=9 ⇔` \(\left[ \begin{array}{l}x=3\ (TM)\\x=-3\ (L)\end{array} \right.\) 

Vậy `S={3}`

Câu 2:

`x^2+\frac{1}{x-1}=1+\frac{1}{x-1}`

ĐK: `x \ne 1`

`⇔ \frac{x^2(x-1)}{x-1}+\frac{1}{x-1}=\frac{x-1}{x-1}+\frac{1}{x-1}`

`⇔ \frac{x^3-x^2+1}{x-1}=\frac{x-1+1}{x-1}`

`⇔ x^3-x^2+1=x`

`⇔ x^3-x^2-x+1=0`

`⇔ x^2(x-1)-(x-1)=0`

`⇔ (x^2-1)(x-1)=0`

`⇔ (x-1)^2(x+1)=0`

`⇔` \(\left[ \begin{array}{l}x=1\ (L)\\x=-1\ (TM)\end{array} \right.\) 

Vậy `S={-1}`

1C1C

2B2B

Giải thích các bước giải:

 x4−8x2−9x+3=0x4-8x2-9x+3=0

ĐK: x≠−3x≠-3

⇔x4−8x2−9=0⇔x4-8x2-9=0

Đặt x2=t (t≥0)x2=t (t≥0)

⇔t2−8t−9=0⇔t2-8t-9=0

⇔t2+t−9t−9=0⇔t2+t-9t-9=0

⇔(t+1)(t−9)=0⇔(t+1)(t-9)=0

⇔⇔ [t=−1 (L)t=9 (TM)[t=−1 (L)t=9 (TM) 

+) t=9→x2=9⇔t=9→x2=9⇔ [x=3 (TM)x=−3 (L)[x=3 (TM)x=−3 (L) 

Vậy S={3}S={3}

Câu 2:

x2+1x−1=1+1x−1x2+1x-1=1+1x-1

ĐK: x≠1x≠1

⇔x2(x−1)x−1+1x−1=x−1x−1+1x−1⇔x2(x-1)x-1+1x-1=x-1x-1+1x-1

⇔x3−x2+1x−1=x−1+1x−1⇔x3-x2+1x-1=x-1+1x-1

⇔x3−x2+1=x⇔x3-x2+1=x

⇔x3−x2−x+1=0⇔x3-x2-x+1=0

⇔x2(x−1)−(x−1)=0⇔x2(x-1)-(x-1)=0

⇔(x2−1)(x−1)=0⇔(x2-1)(x-1)=0

⇔(x−1)2(x+1)=0⇔(x-1)2(x+1)=0

⇔⇔ [x=1 (L)x=−1 (TM)[x=1 (L)x=−1 (TM) 

Vậy S={−1}