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a) \(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)
\(=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\frac{\left(x+1\right)\left(x^3+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)
b) Xét tử ta có: \(\left(x+1\right)^2\ge0\) (1)
Xét mấu ta có: \(x^2\ge0\Rightarrow x^2+1\ge1>0\) (2)
Từ (1) và (2) \(\Rightarrow\) Phân thức trên k âm với mọi x
Mình làm tắt thôi nhé
\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x^2-x+1\right)}=\frac{\left(x-1\right)^2}{x^2-x+1}\left(x\ne-1\right)\)
Dễ thấy \(A\ge0\)
\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1}{x^4-x^3+x^2+2x^2-2x^2+2x+x^2-x+1}\)
\(=\frac{x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)}{x^2\left(x^2-x+1\right)+2x\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x^2-2x+1\right)}{\left(x^2+2x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^2-2x+1}{x^2-x+1}\)
\(=\frac{\left(x-1\right)^2}{x^2-x+1}\)
Ta có : \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\)
=> Đpcm
\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
\(x^4-x^3+2x^2-x+1=\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(x^2+1\right)\)
Ta có: \(\left(x+1\right)^2\ge0;\forall x\)
\(x^2+1>1\); \(\forall x\)
\(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0,\forall x\)
Vậy \(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{\left(x+1\right)^2}{x^2+1}\ge0;\forall x\)
\(a,Đkxđ:x\ne\pm2\)
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+1\right)^2}{x^2-4}\)
b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)
Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)
\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)
Vậy ............
\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)
Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0