\(\frac{x^4-5x^2+4}{x^4-10x^2+9}\)    \(ĐKXĐ:x\ne\pm3\)

\(=\frac{x^4-4x^2-x^2+4}{x^4-9x^2-x^2+9}\)

\(=\frac{\left(x^4-4x^2\right)-\left(x^2-4\right)}{\left(x^4-9x^2\right)-\left(x^2-9\right)}\)

\(=\frac{x^2.\left(x^2-4\right)-\left(x^2-4\right)}{x^2.\left(x^2-9\right)-\left(x^2-9\right)}\)

\(=\frac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)

\(=\frac{x^2-4}{x^2-9}\)    

\(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\frac{x^2.\left(x^2-1\right)-4.\left(x^2-1\right)}{x^2.\left(x^2-1\right)-9.\left(x^2-1\right)}\)

                                                                           \(=\frac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\frac{x^2-4}{x^2-9}\)

\(A=\frac{x^4-5x^2+4}{x^4-10^2+9}=\frac{x^2\left(x^2-5+4\right)}{x^2\left(x^2-10+9\right)}\)

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

sai r bn ơi .mik lm đc r

`1)` Biểu thức xác định `<=>x+1 \ne 0<=>x \ne -1`

`[x^2+2x+1]/[x+1]=[(x+1)^2]/[x+1]=x+1`

`2)` Bth xác định `<=>x(x-3) \ne 0<=>{(x \ne 0),(x \ne 3):}`

`[x^2-6x+9]/[x(x-3)]=[(x-3)^]/[x(x-3)]=[x-3]/x`

`3)` Bth xác định `<=>2x(x+2) \ne 0<=>{(x \ne 0),(x \ne -2):}`

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

`4)` Bth xác định `<=>5x^2-10x \ne 0<=>5x(x-2) \ne 0<=>{(x \ne 0),(x \ne 2):}`

`[x^2-2x]/[5x^2-10x]=[x(x-2)]/[5x(x-2)]=1/5`

1)

\(ĐKXĐ:x\ne-1\)

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

2)

ĐKXĐ x khác 0 và x khác 3

\(\dfrac{x^2-6x+9}{x\left(x-3\right)}\\ =\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\\ =\dfrac{x-3}{x}\)

3)

ĐKXĐ: x khác 0 và x khác -2

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

4)

DKXĐ: x khác 0 và x khác 2

\(\dfrac{x^2-2x}{5x^2-10x}\\ =\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}\\ =\dfrac{1}{5}\)

đk `x≠-1`

`(x^2+2x+1)/(x+1)`

`=((x+1)^2)/(x+1)`

`=x+1`

---------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

`(x^2-6x+9)/(x(x-3))`

`=((x-3)^2)/(x(x-3))`

`=(x-3)/x`

--------

 đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)

`(x^2-4)/(2x(x+2))`

`=((x-2)(x+2))/(2x(x+2))`

`=(x-2)/(2x)`

--------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

`(x^2-2x)/(5x^2-10x)`

`=(x(x-2))/(5x(x-2))`

`=x/(5x)`

\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)

ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)

b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)

a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

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

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

\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)

b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)

\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)

=0

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

b) Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: Để B=0 thì x=1

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:

\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Vậy: Khi x=3 thì B=1

d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: 

\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Để B>0 thì \(\dfrac{x-1}{2}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để B>0 thì x>1