a) x ≠ -5.

b) Ta có P = ( x + 5 ) 2 x + 5 = x + 5  

c) Ta có P = 1 Û x = -4 (TMĐK)

d) Ta có P = 0 Û x = -5 (loại). Do vậy x ∈ ∅ .

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

\(A=\dfrac{x-1}{x+1}\)

c) Thay x = 3 vào A ta có:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

\(B=\dfrac{2\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}\)

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

đề bài có sai ko zậy                

k sai đâu bn

1.Cho  \(\frac{x^2-4x+4}{x^2-4}< 2\)

<=>\(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}< 2\)

<=>\(\frac{x-2}{x+2}< 2\)

<=>\(\frac{x-2}{x+2}-2< 0\)

<=>\(\frac{x-2}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)

<=>\(\frac{x-2-2\left(x+2\right)}{x+2}< 0\)

<=>\(\frac{x-2-2x-4}{x+2}< 0\)

<=>\(\frac{-x-6}{x+2}< 0\)

<=>\(\orbr{\begin{cases}\hept{\begin{cases}-x-6< 0\\x+2>0\end{cases}}\\\hept{\begin{cases}-x-6>0\\x+2< 0\end{cases}}\end{cases}}\)

<=>\(\orbr{\begin{cases}\hept{\begin{cases}x< -6\\x< -2\end{cases}}\\\hept{\begin{cases}x>-6\\x>-2\end{cases}}\end{cases}}\)

<=>\(\orbr{\begin{cases}x< -2\\x>-6\end{cases}}\)

Vậy -6 < x < -2

1) \(\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

\(\frac{x-2}{x+2}< 2\)

\(\Leftrightarrow\frac{x-2}{x+2}-2< 0\)

\(\Leftrightarrow\frac{x-2}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)

\(\Leftrightarrow\frac{x-2-2x-4}{x+2}< 0\)

\(\Leftrightarrow\frac{-x-6}{x+2}< 0\)

\(\Leftrightarrow-x-6< 0\)

\(\Leftrightarrow-x< 6\)

\(\Leftrightarrow x>-6\)

vậy \(x>-6\)thì giá trị của phân thức \(>2\)

2) \(\frac{2x^2-4x+8}{x^3+8}\)

\(=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(=\frac{2}{x+2}\)\(\left(x\ne-2\right)\)

khi đó \(\frac{2}{x+2}>2\)

\(\Leftrightarrow\frac{2}{x+2}-2>0\)

\(\Leftrightarrow\frac{2}{x+2}-\frac{2\left(x+2\right)}{x+2}>0\)

\(\Leftrightarrow\frac{2-2x-4}{x+2}>0\)

\(\Leftrightarrow\frac{-2x-2}{x+2}>0\)

\(\Leftrightarrow-2x-2>0\)

\(\Leftrightarrow-2x>2\)

\(\Leftrightarrow x< -1\)

1) Ta có: 
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz 
= [(x+y)³ + z³] - 3xy(x+y+z) 
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z) 
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy] 
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy) 
= (x+y+z)(x² + y² + z² - xy - xz - yz). 

Câu 2:

\(\frac{x^2-y^2+6x+9}{x+y+3}\)

\(=\frac{x^2-y^2+x^2+6x+9-x^2}{x+y+3}\)

\(=\frac{ \left(x+3\right)^2-y^2}{x+y+3}\)

\(=\frac{\left(x-y+3\right)\left(x+y+3\right)}{x+y+3}\)

\(=x-y+3\)

`a)ĐKXĐ` của `D` la `x+2 \ne 0<=>x \ne -2`

`b)` Với `x \ne -2` có: `D=[2x^2-4x+8]/[x^3+8]`

                                `D=[2(x^2-2x+4)]/[(x+2)(x^2-2x+4)]=2/[x+2]`

`c)` Thay `x=2` vào `D` có: `D=2/[2+2]=1/2`

`d)D > 2<=>2/[x+2] > 2`

         `<=>[2-2x-4]/[x+2] > 0`

         `<=>[x+1]/[x+2] < 0<=>-2 < x <= -1`

a: ĐKXĐ: \(x\notin\left\{-1;-2\right\}\)

b: \(M=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\dfrac{x-1}{x+2}\)

Thay x=2002 vào M, ta được:

\(M=\dfrac{2002-1}{2002+1}=\dfrac{2001}{2003}\)

c: Để M=0 thì x-1=0

hay x=1(nhận)