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Để : \(\frac{3}{x+1}\in Z\) thì 3 chia hết cho n + 1
=> n + 1 thuộc Ư(3) = {-3;-1;1;3}
Ta có bảng ;
| n + 1 | -3 | -1 | 1 | 3 |
| n | -4 | -2 | 0 | 2 |
cảm ơn bạn nha # Nguyễn Việt Hoàng
bạn giúp mik những câu sau được không
`a)C=((2x^2+1)/(x^3-1)-1/(x-1)):(1-(x^2-2)/(x^2+x+1))`
`ĐK:x ne 1`
`C=((2x^2+1-x^2-x-1)/(x^3-1)):((x^2+x+1-x^2+2)/(x^2+x+1))`
`C=((x^2-x)/(x^3-1)):((x+3)/(x^2+x+1))`
`C=x/(x^2+x+1)*(x^2+x+1)/(x+3)`
`C=x/(x+3)`
`b)|1-x|+2=3(x+1)`
`<=>|1-x|+2=3x+3`
`<=>|1-x|=3x+1(x>=-1/3)`
`**1-x=3x+1`
`<=>4x=0<=>x=0(tmđk)`
`**x-1=3x+1`
`<=>2x=-2`
`<=>x=-1(l)`
Thay `x=0` vào C
`=>C=0`
`c)C in ZZ`
`=>x vdots x+3`
`=>x+3-3 vdots x+3`
`=>3 vdots x+3`
`=>x+3 in Ư(3)={+-1,+-3}`
`=>x in {-2,-4,0,-6}`
`d)|C|>C`
Mà `|C|>=0`
`=>C<0`
`<=>x/(x+3)<0`
Để 1 p/s `<=0` thì tử và mẫu trái dấu mà `x<x+3`
`=>` \(\begin{cases}x<0\\x+3>0\\\end{cases}\)
`<=>` \(\begin{cases}x>-3\\x<0\\\end{cases}\)
`<=>-3<x<0`
a: A nguyên
=>3x+1 chia hết cho 2-x
=>3x-6+7 chia hết cho x-2
=>x-2 thuộc {1;-1;7;-7}
=>x thuộc {3;1;9;-5}
b: B nguyên
=>8x-4+6 chia hết cho 2x-1
=>2x-1 thuộc {1;-1;2;-2;3;-3;6;-6}
=>x thuộc {1;0;2;-1}
c: C nguyên
=>x-1 chia hết cho 2x+1
=>2x-2 chia hết cho 2x+1
=>2x+1-3 chia hết cho 2x+1
=>2x+1 thuộc {1;-1;3;-3}
=>x thuộc {0;-1;1;-2}
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)