cho bieu thuc \(Q=\frac{12x-15}{x^2-7x+12}-\frac{x+5}{x-4}+\frac{2x-3}{3-x}\) . Tim x de gia tri cua Q nguyen
K
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KN
10 tháng 2 2020
a) \(A< 0\Leftrightarrow\frac{x^2+3}{x-2}< 0\)
Mà \(x^2+3>0\Rightarrow x-2< 0\Leftrightarrow x< 2\)
b) \(A\inℤ\Leftrightarrow\frac{x^2+3}{x-2}\in Z\)
Ta có \(\frac{x^2+3}{x-2}=\frac{\left(x^2-4x+4\right)+\left(4x-8\right)+7}{x-2}\)
\(=x-2+4+\frac{7}{x-2}\)
\(\Rightarrow\frac{x^2+3}{x-2}\in Z\Leftrightarrow7⋮\left(x-2\right)\)
\(\Rightarrow x-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{3;1;9;-5\right\}\)
9 tháng 1 2018
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 ^^
9 tháng 1 2018
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;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
H
14 tháng 12 2018
\(A=\frac{3}{2-x}+\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(A=\frac{-3}{x-2}+\frac{3}{x+2}+\frac{3x^2}{\left(x+2\right)\left(x-2\right)}\)
\(A=\frac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-3x-6+3x-6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-12+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(-4+x^2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(A=3\)
14 tháng 12 2018
\(a,A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(=\frac{-3\left(x+2\right)-3\left(x-2\right)+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x^2-6x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x}{x+2}\)
\(b,ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x+2\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm2\\x\ne-1\end{cases}}}\)
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 :
| x + 1 | -3 | -1 | 1 | 3 |
| x | -4 | -2 | 0 | 2 |
Vậy \(x\in\left\{-4;-2;0;2\right\}\)