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dkxd \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)
b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)
\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)
tới khúc này bí rồi ^^
a,ĐKXĐ của A là:\(x\ne+2;-2\)
b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)
c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)
Ta có bảng
| \(x^2-4\) | x |
| 4 | \(\sqrt{8}\) |
| -4 | 0 |
| 2 | \(\sqrt{6}\) |
| -2 | \(\sqrt{2}\) |
| 1 | \(\sqrt{5}\) |
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
a: ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)
b: \(B=\dfrac{x+3-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+2+1}{x+2}\)
\(=\dfrac{x+2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+2}=\dfrac{1}{x-3}\)
c: Để B nguyên thì \(x-3\in\left\{1;-1\right\}\)
hay \(x\in\left\{4;2\right\}\)
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(A=\frac{3}{2-x}+\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(A=\frac{-3}{x-2}+\frac{3}{x+2}+\frac{3x^2}{\left(x+2\right)\left(x-2\right)}\)
\(A=\frac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-3x-6+3x-6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-12+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(-4+x^2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(A=3\)
\(a,A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(=\frac{-3\left(x+2\right)-3\left(x-2\right)+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x^2-6x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x}{x+2}\)
\(b,ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm2\\x\ne-1\end{cases}}}\)
Ta có : \(P=A:B=\frac{3x}{x+2}:\frac{x+1}{x+2}\)
\(=\frac{3x}{x+2}.\frac{x+2}{x+1}\)
\(=\frac{3x}{x+1}\)
\(=\frac{3x+3}{x+1}-\frac{3}{x+1}\)
\(=3-\frac{3}{x+1}\)
Để P nguyên thì \(3-\frac{3}{x+1}\inℤ\)
\(\Leftrightarrow\frac{3}{x+1}\inℤ\)
Vì \(x\inℤ\Rightarrow x+1\inℤ\)
Ta có bảng :
Vậy \(x\in\left\{-4;-2;0;2\right\}\)