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a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)
Ta có : \(P=\dfrac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{4}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x\left(x-3\right)+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x-x+6}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)
b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)
- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)
Vậy ...
a ĐKXĐ : \(x\ne2,x\ne3\)
\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)
Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)
\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\)
Thử lại ta thấy đúng
Vậy...
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
\(a,A=\dfrac{x\left(x+2\right)+\left(2-x\right)\left(x-2\right)+12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2x-4-x^2+2x+12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-4x+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{4}{x+2}\)
Vậy \(A=-\dfrac{4}{\left(x+2\right)}\)
để A xác định
\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)
\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)
\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)
\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)
C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)
d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)
\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)
\(x^2-4\inℕ\)
\(\Rightarrow x^2\in\left\{0;4;12\right\}\)
Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn
mk học lớp 6 mong mn thông cảm nếu có sai sót
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)
ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)
b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)