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a) Ta có: \(P=\left(\dfrac{x^2-1}{x^4-x^2+1}+\dfrac{2}{x^6+1}-\dfrac{1}{x^2+1}\right)\cdot\left(x^2-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)

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

\(=\dfrac{x^4-1+2-x^4+x^2-1}{\left(x^2+1\right)\cdot\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+x^4+x^2-x^4-x^2+1}{x^4+x^2+1}\)

\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+1}{x^4+x^2+1}\)

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

\(=\dfrac{x^2}{x^4+x^2+1}\)

18 tháng 2 2021

giúp e phần b với ạ

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị

17 tháng 4 2022

B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)

\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)

b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)

\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))

\(\Leftrightarrow x>-1\).

-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).

 

a: ĐKXĐ: x<>1; x<>2; x<>3

\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)

\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)

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

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

b:

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AH
Akai Haruma
Giáo viên
19 tháng 4 2021

Lời giải:
ĐK: $x\neq \pm 2; x\neq 0$

a) 

\(A=\left[\frac{x+2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}\right].\frac{2-x}{x}=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}\)

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

b) Để $A=1\Leftrightarrow \frac{-4}{x+2}=1$

$\Leftrightarrow x+2=-4$

$\Leftrightarrow x=-6$ (thỏa ĐKXĐ)

Vậy $x=-6$

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

a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)

\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)

\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)

\(=\dfrac{3x}{x-2}\)

b) Để A nguyên thì \(3x⋮x-2\)

\(\Leftrightarrow3x-6+6⋮x-2\)

mà \(3x-6⋮x-2\)

nên \(6⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(6\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{3;1;4;0;5;8;-4\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)