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Điều kiện xác định của \(P\)là:
\(\hept{\begin{cases}x^2+2x+1\ne0\\x^2-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)
\(P=\left(\frac{2+x}{x^2+2x+1}-\frac{x-2}{x^2-1}\right).\frac{1-x^2}{x}\)
\(=\left[\frac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}-\frac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\frac{1-x^2}{x}\)
\(=\frac{2x}{\left(x+1\right)^2\left(x-1\right)}.\frac{1-x^2}{x}=\frac{-2}{x+1}\)
Để \(P\)nguyên mà \(x\)nguyên suy ra \(x+1\inƯ\left(2\right)=\left\{-2,-1,1,2\right\}\Leftrightarrow x\in\left\{-3,-2,0,1\right\}\)
Đối chiếu điều kiện ta được \(x\in\left\{-3,-2\right\}\)thỏa mãn.
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
a/. ĐKXĐ : (x-1)(x+1) # 0 => x # 1 hay x # -1
b/. \(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3.2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-4x-3}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{2\left(4-2x\right)}{5}\)
Em xem lại đè nhé. Đề như vậy thì sẽ ko rút gọn đc hết x trên tử. nên B vẫn phụ thuộc vào biến x.
chao cac bạn và a chi nếu đề sửa lai vây thi minh làm thế nào ( x+1/2x-2 + 3/x^2+1 - x+3/2x+1 )* (4x^2 -1)/5
a) M xác định khi \(x+1\ne0\)
\(x^2+1\ne0\)
\(x^2+2x+1=\left(x+1\right)^2\ne0\)
\(\Leftrightarrow x\ne\pm1\)
b) \(M=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{1}{x^2+2x+1}-\frac{1}{x^2-1}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{1}{\left(x+1\right)^2}-\frac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{1\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}-\frac{1\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{\left[1\left(x^2-1\right)\right]-1\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}.\frac{x^2-1-1\left(x^2+2x+1\right)}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}.\frac{x^2-1-x^2-2x-1}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}.\frac{-2x-2}{\left(x+1\right)^2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{1}{x+1}+\frac{\left(x-x^3\right)\left(-2x-2\right)}{\left(x^2+1\right)\left(x^2-1\right)\left(x+1\right)^2}\)\(=\frac{1}{x+1}+\frac{\left(x-x^3\right)\left(-2x-2\right)}{\left(x^4-1\right)\left(x+1\right)^2}\)
\(=\frac{1}{x+1}+\frac{-2\left(x-x^3\right)\left(x+1\right)}{\left(x^4-1\right)\left(x+1\right)^2}\)\(=\frac{1}{x+1}+\frac{-2\left(x-x^3\right)}{\left(x^4-1\right)\left(x+1\right)}\)
\(=\frac{\left(x^4-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^4-1\right)\left(x+1\right)}+\frac{-2\left(x-x^3\right)\left(x+1\right)}{\left(x^4-1\right)\left(x+1\right)}\)
\(=\frac{\left(x^4-1\right)}{\left(x+1\right)\left(x^4-1\right)}+\frac{-2\left(x-x^3\right)}{\left(x^4-1\right)}\)\(=\frac{1}{x+1}+\frac{-2\left(x-x^3\right)}{\left(x^4-1\right)}\)??? Chắc hết rút được rồi :v
Câu b) hơi dài quá rồi.Làm lại
b) \(M=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{1}{x^2+2x+1}-\frac{1}{x^2-1}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{1}{\left(x+1\right)^2}-\frac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{x-1}{\left(x+1\right)^2\left(x-1\right)}-\frac{x+1}{\left(x+1\right)^2\left(x-1\right)}\right)\)
\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}\left(\frac{\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\right)\)\(=\frac{1}{x+1}+\frac{x-x^3}{x^2+1}.\frac{-2}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\frac{1}{x+1}+\frac{-2\left(x-x^3\right)}{\left(x^2+1\right)\left(x+1\right)^2\left(x-1\right)}\)\(=\frac{1}{x+1}+\frac{2x\left(x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)^2\left(x-1\right)}\)
\(=\frac{1}{x+1}+\frac{2x}{\left(x^2+1\right)\left(x+1\right)}=\frac{x+1}{x^2+1}\) (Quy đồng và rút gọn)