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\(a,A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2-4}\)
Vậy \(A=\dfrac{\left(x+1\right)^2}{x^2-4}\)
\(b,\) Theo đề, ta có : \(-2< x< 2\)
\(\Rightarrow x-2< 0;x+2>0;\left(x+1\right)^2>0\)
\(\Rightarrow A< 0\) hay phân thức luôn có giá trị âm
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
a) Ta có: \(C=\left(x^2-1\right)\cdot\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}+1\right)\)
\(=\left(x^2-1\right)\cdot\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\left(x^2-1\right)\cdot\dfrac{x+1-x+1+x^2-1}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2-1\)
a) \(A= \dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4} \\ =\dfrac{1}{x-2}+\dfrac{1}{x-2}+\dfrac{x^2+1}{(x-2)(x+2)} \\= \dfrac{x+2+x-2+x^2+1}{(x-2)(x+2)} \\=\dfrac{x^2+2x+1}{x^2-4} \\ =\dfrac{(x+1)^2}{(x-2)(x+2)}\)
b) Với mọi \(x\) thỏa mãn \(-2<x<2\) và \(x \ne -1\) thì \(x-2\) đều có giá trị âm, mà \(\begin{cases}(x+1)^2≥0\\x+2>0\\\end{cases}\) \( \Rightarrow\) Biểu thức A luôn có giá trị âm.
để 
nhận giá trị nguyên
với 
Tìm tất cả các số tự nhiên 
để 
là số nguyên tố.
Câu 1:
a) \(A=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}.\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}\)
\(=\left[\dfrac{2}{3x}-\dfrac{2}{3x}+\dfrac{2x}{x+1}+\dfrac{2}{x+1}\right]\dfrac{x}{x-1}\)
\(=\left[\dfrac{2x}{x+1}+\dfrac{2}{x+1}\right]\dfrac{x}{x-1}\)
\(=\dfrac{2x+2}{x+1}.\dfrac{x}{x-1}\)
\(=\dfrac{2\left(x+1\right)}{x+1}.\dfrac{x}{x-1}\)
\(=2.\dfrac{x}{x-1}\)
\(=\dfrac{2x}{x-1}\)
Câu 1:
ĐKXĐ: \(x\notin\left\{0;-1;1\right\}\)
a) Ta có: \(A=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\left(\dfrac{x+1}{3x}-x-1\right)\right):\dfrac{x-1}{x}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\left(\dfrac{x+1}{3x}-\dfrac{3x\left(x+1\right)}{3x}\right)\right):\dfrac{x-1}{x}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right):\dfrac{x-1}{x}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right):\dfrac{x-1}{x}\)
\(=\left(\dfrac{2\left(x+1\right)}{3x\left(x+1\right)}-\dfrac{2\cdot\left(-3x^2-2x+1\right)}{3x\left(x+1\right)}\right):\dfrac{x-1}{x}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)
\(=\dfrac{6x^2+6x}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)
\(=\dfrac{6x\left(x+1\right)}{3x\left(x+1\right)}:\dfrac{x-1}{x}\)
\(=2\cdot\dfrac{x}{x-1}=\dfrac{2x}{x-1}\)
b) Để A nguyên thì \(2x⋮x-1\)
\(\Leftrightarrow2x-2+2⋮x-1\)
mà \(2x-2⋮x-1\)
nên \(2⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(2\right)\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;3\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{2;3\right\}\)
\(a,B=4x^2+20x+25-9+x^2+14=5x^2+20x+30\\ b,B=5\left(x^2+4x+4\right)+10\\ B=5\left(x+2\right)^2+10\ge10>0,\forall x\)
Do đó B luôn dương với mọi x
a) \(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(A=4x^2+4x+1-x^2+4-2x^2-2x\)
\(A=x^2+2x+5\)
b) Để A = 4
=> \(x^2+2x+5=4\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c) Ta có A = x2 + 2x + 5
A = ( x + 1 )2 + 4
=> \(A\ge4>0\left(đpcm\right)\)
a,\(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(=4x^2+4x+1-x^2+4-2x^2-2x\)
\(=x^2+2x+5\)
b,\(A=x^2+2x+5=4\)
\(\Rightarrow x^2+2x+5-4=0\)
\(x^2+2x+1=0\)
\(\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
c, Ta có: \(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4>0\)
Hay: A > 0 => đpcm
=.= hok tốt!!
Biểu thức đâu bạn :V