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\(B=\frac{2x^2-2}{x^3+x^2-x-1}=\frac{2\left(x-1\right)\left(x+1\right)}{x^2\left(x+1\right)-\left(x+1\right)}=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)^2}\)
\(ĐKXĐ:x\ne\pm1\)(1)
\(\)\(B=\frac{2}{x+1}\)
Để B thuộc Z => \(2⋮x+1\left(x\in Z\right)\)
\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left(1;-1;2;-2\right)\)
\(\Rightarrow x\in\left(0;-2;1;-3\right)\)(2)
từ (1) và (2)
\(\Rightarrow x\in\left(0;-2;-3\right)\)
a: Sửa đề: \(A=\dfrac{x^3+2x^2+6x+8}{x+1}\)
Để A là số nguyên thì \(x^3+x^2+x^2+x+5x+5+3⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{0;-2;2;-4\right\}\)
b: Để \(\dfrac{2x^2+x-2}{x-3}\) là số nguyên thì \(2x^2-6x+7x-21+19⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;19;-19\right\}\)
hay \(x\in\left\{4;2;22;-16\right\}\)
a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)
b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;51;6;0;9;-3\right\}\)
c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)
=>2x^2-2x+6x-6=x-3
=>2x^2+5x-6-x+3=0
=>2x^2+4x-3=0
hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{x^3-1}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1}{x-1}\)
Vậy \(A=\dfrac{x+1}{x-1}\)
Giả sử tìm được \(x\in Z\) để \(A\in Z\)
\(x\in Z\Leftrightarrow\left\{{}\begin{matrix}x+1\in Z\\x-1\in Z\end{matrix}\right.\)
\(A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)
\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)
Ta có các trường hợp :
+) \(x-1=1\Leftrightarrow x=2\)
+) \(x-1=2\Leftrightarrow x=3\)
+) \(x-1=-1\Leftrightarrow x=0\)
+) \(x-1=-2\Leftrightarrow x=-1\)
Vậy..
a) \(E=\left(\frac{1}{x+2}+\frac{1}{x-2}\right).\frac{x-2}{x}\left(ĐKXĐ:x\ne0;x\ne\pm2\right)\)
\(=\left(\frac{x-2+x+2}{\left(x+2\right)\left(x-2\right)}\right).\frac{x-2}{x}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x}=\frac{2x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=\frac{2}{x+2}\)
b) Khi x = 6 \(\Rightarrow E=\frac{2}{x+2}=\frac{2}{6+2}=\frac{2}{8}=\frac{1}{4}\)
c) \(E=4\Leftrightarrow\frac{2}{x+2}=4\Leftrightarrow4\left(x+2\right)=2\Leftrightarrow4x+8=2\Leftrightarrow x=\frac{-3}{2}\)
Vậy để E = 4 thì x = -3/2
d) \(E>0\Leftrightarrow\frac{2}{x+2}>0\Leftrightarrow2>0\)
Vậy phương trình vô nghiệm
e) \(E\in Z\Leftrightarrow x+2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Nếu x + 2 = 1 thì x = -1
Nếu x + 2 = -1 thì x = -3
Nếu x + 2 = 2 thì x = 0
Nếu x + 2 = -2 thì x = -4
Vậy ...
Nek bạn giải thích hộ mik tí nữa nhé :Tại sao 2 > 0 thì phương trình lại vô nghiệm ?
