Đáp án + Giải thích các bước giải:

Câu `10` :

`a)` Ta có : `4x^2=1=>x^2=1/4=>x=1/2` ( vì `xne -1/2` )

Thay `x = 1/2` vào biểu thức `B` ta có :

`B=(x^2-x)/(2x+1)=[(1/2)^2-1/2]/[2*1/2+1]=(1/4-1/2)/2=(-1/4):2=-1/8`

Vậy `B=-1/8`

`b)` 

+ Rút gọn biểu thức `A` :

`A=1/(x-1) - x/(1-x^2)=1/(x-1)+x/(x^2-1)` $\\$ `= (x+1)/[(x-1)(x+1)]+x/[(x-1)(x+1)]=(x+1+x)/[(x-1)(x+1)]=(2x+1)/[(x-1)(x+1)]`

`M=A*B=(2x+1)/[(x-1)(x+1)]*[x(x-1)]/(2x+1)=x/(x+1)`

`c)` `x/(x+1)<1<=>x/(x+1)-1<0 <=>x/(x+1)-(x+1)/(x+1)<0` $\\$ `<=> (x-x-1)/(x+1)<0<=>(-1)/(x+1)<0<=>x+1>0` ( vì `x ne -1` )

`<=>x> -1`

Câu `11` :

`a)` `|x-1|=2 => `\(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\) $\\$ `=> `\(\left[ \begin{array}{l}x=3(tm)\\x=-1(ktm)\end{array} \right.\)

Thay `x = 3` vào biểu thức `A` ta có :

`A=(x^2-2x)/(x+1)=(3^2-2*3)/(3+1)=(9-6)/4=3/4`

Vậy `A=3/4`

`b)`

+ Rút gọn `B` :

`B=(x+2)/(x-2)-(x-2)/(x+2)-16/(4-x^2)` $\\$ `= (x+2)/(x-2) - (x-2)/(x+2)+16/(x^2-4)` $\\$ `= [(x+2)(x+2)]/[(x-2)(x+2)]-[(x-2)(x-2)]/[(x-2)(x+2)]+16/[(x-2)(x-2)]` $\\$ `= (x^2+4x+4-(x^2-4x+4)+16)/[(x-2)(x+2)]` $\\$ `= (x^2 + 4x + 4 - x^2 + 4x - 4 + 16)/[(x-2)(x+2)]` $\\$ `= (8x + 16)/[(x-2)(x+2)]=[8(x+2)]/[(x-2)(x+2)]=8/(x-2)`

+ Rút gọn `P` :

`P=A*B=(x^2-2x)/(x+1)*8/(x-2)=[x(x-2)]/(x+1)*8/(x-2)=(8x)/(x+1)`

`c)` 

`P<8<=>(8x)/(x+1)<8<=>(8x)/(x+1)-8<0` $\\$ `<=>(8x)/(x+1)-[8(x+1)]/(x+1)<0 ` $\\$ `<=>(8x-8x-8)/(x+1)<0` $\\$ `<=> (-8)/(x + 1)<0`

Vì `-8<0<=>x+1>0<=>x> -1` ( do `x ne -1`)

10

a/ ĐKXĐ: $x\ne -\dfrac{1}{2}$

$4x^2=1\\↔x^2=\dfrac{1}{4}\\↔x=±\dfrac{1}{2}$

mà $x\ne -\dfrac{1}{2}$

$→x=\dfrac{1}{2}$

Với $x=\dfrac{1}{2}$ (thỏa mãn điều kiện)

$→B=\dfrac{(\dfrac{1}{2})^2-\dfrac 1 2}{2.\dfrac 1 2 +1}=-\dfrac{1}{8}$

b/ ĐKXĐ: $x\ne -1;1;-\dfrac{1}{2}$

$M=A.B\\=\left(\dfrac{1}{x-1}-\dfrac{x}{1-x^2}\right).\dfrac{x^2-x}{2x+1}\\=\left(\dfrac{x+1}{(x-1)(x+1)}+\dfrac{x}{x^2-1}\right).\dfrac{x(x-1)}{2x+1}\\=\left(\dfrac{x+1}{(x-1)(x+1)}+\dfrac{x}{(x-1)(x+1)}\right).\dfrac{x(x-1)}{2x+1}\\=\dfrac{x+1+x}{(x-1)(x+1)}.\dfrac{x(x-1)}{2x+1}\\=\dfrac{2x+1}{x+1}.\dfrac{x}{2x+1}\\=\dfrac{x}{x+1}$

c/ $M<1\\↔\dfrac{x}{x+1}<1\\↔\dfrac{x}{x+1}-1<0\\↔\dfrac{x-x-1}{x+1}<0\\↔\dfrac{-1}{x+1}<0\\↔x+1>0\\↔x>-1$

Vậy $x>-1$ thì $M<1$

11

a/

$|x-1|=2\\↔\left[\begin{array}{1}x-1=2\\x-1=-2\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=-1\end{array}\right.$

mà $x\ne -1$

$→x=3$ 

Với $x=3$ (thỏa mãn điều kiện)

$→A=\dfrac{3^2-2.3}{3+1}=\dfrac{3}{4}$

b/ $P=A.B\\=\dfrac{x^2-2x}{x+1}.\left(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}-\dfrac{16}{4-x^2}\right)\\=\dfrac{x(x-2)}{x+1}.\left(\dfrac{(x+2)^2}{(x-2)(x+2)}-\dfrac{(x-2)^2}{(x-2)(x+2)}+\dfrac{16}{x^2-4}\right)\\=\dfrac{x(x-2)}{x+1}.\left(\dfrac{x^2+4x+4-x^2+4x-4}{(x-2)(x+2)}+\dfrac{16}{(x-2)(x+2)}\right)\\=\dfrac{x(x-2)}{x+1}.\dfrac{8x+16}{(x-2)(x+2)}\\=\dfrac{x}{x+1}.\dfrac{8(x+2)}{x+2}\\=\dfrac{8x}{x+1}$

c/ $P<8\\↔\dfrac{8x}{x+1}<8\\↔\dfrac{8x}{x+1}-8<0\\↔\dfrac{8x-8x-8}{x+1}<0\\↔\dfrac{-8}{x+1}<0\\↔x+1>0\\↔x>-1$

Vậy $x>-1$ thì $P<8$