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24 tháng 12 2017
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12 tháng 5 2022
\(A=\dfrac{3x^2-9x+x-3+2}{x-3}\)
\(B=\dfrac{x^2\left(x+2\right)+5\left(x+2\right)}{\left(x+2\right)^2}=\dfrac{x^2+5}{x+2}=x-2+\dfrac{9}{x+2}\)
Để A và B cùng là số nguyên thì
\(\left\{{}\begin{matrix}x-3\in\left\{1;-1;2;-2\right\}\\x+2\in\left\{1;-1;3;-3;9;-9\right\}\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\in\left\{4;2;5;1\right\}\\x\in\left\{-1;-3;1;-5;7;-11\right\}\end{matrix}\right.\)
hay x=1
AG
24 tháng 6 2019
Ta có : Để M=\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\left(\frac{x^2+8x+16}{32}\right)=0\)
<=> M=\(\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)=0\)
<=>M=\(\left(\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)
<=>M=\(\left(\frac{32}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)
<=>M=\(\frac{x+4}{x-4}\)
b) Thay x=\(\frac{-3}{8}\) vào M:
M=\(\frac{x+4}{x-4}=\frac{\frac{-3}{8}+4}{\frac{-3}{8}-4}=\frac{-29}{35}\)
c)Hình như sai!
d)
10 tháng 6 2018
a) ta có: \(A=\frac{2x}{x-2}=\frac{2x-4+4}{x-2}=\frac{2.\left(x-2\right)+4}{x-2}=\frac{2.\left(x-2\right)}{x-2}+\frac{4}{x-2}=2+\frac{4}{x-2}\)
Để \(A\inℤ\)
\(\Rightarrow\frac{4}{x-2}\inℤ\)
\(\Rightarrow4⋮x-2\Rightarrow x-2\inƯ_{\left(4\right)}=\left(4;-4;2;-2;1;-1\right)\)
nếu x -2 = 4 => x = 6 (TM)
x- 2= - 4 => x= - 2 (TM)
x- 2= 2 => x = 4 (TM)
x- 2 = -2 => x = 0 (TM)
x - 2 = 1 => x = 3 (TM)
x - 2 = -1 => x= 1 (TM)
KL: \(x\in\left(6;-2;4;0;3;1\right)\)
c) ta có: \(C=\frac{x^2+2}{x+1}=\frac{\left(x+1\right).\left(x-1\right)+3}{x+1}=\frac{\left(x+1\right).\left(x-1\right)}{x+1}+\frac{3}{x+1}\)\(=x-1+\frac{3}{x+1}\)
Để \(C\inℤ\)
\(\Rightarrow\frac{3}{x+1}\inℤ\)
\(\Rightarrow3⋮x+1\Rightarrow x+1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)
nếu x + 1 = 3 => x = 2 (TM)
x + 1 = - 3 => x = -4 (TM)
x + 1 = 1 => x = 0
x + 1 = -1 => x = -2 (TM)
KL: \(x\in\left(2;-4;0;-2\right)\)
p/s
13 tháng 6 2016
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)
\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\)\(\frac{x^2-3x-1}{3x}\)
\(\frac{\left(1-2x\right)\left(1+2x\right)}{3x\left(1-2x\right)}+\frac{x^2-3x-1}{3x}=\frac{x^2-3x-1+1+2x}{3x}=\)\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
13 tháng 6 2016
a)\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)
\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}\)
\(\frac{\left(1-2x\right)\left(1+2x\right)}{3x\left(1-2x\right)}+\frac{x^2-3x-1}{3x}=\frac{x^2-3x-1+1+2x}{3x}=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b) \(\left|x\right|=\frac{1}{3}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(x\ge0\right)\\x=-\frac{1}{3}\left(x< 0\right)\end{cases}}\)
Thay vào \(\frac{x-1}{3}\)tính được A.
c) \(A< 0\Rightarrow\frac{x-1}{3}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Kết hợp cùng với điều kiện của ở phần rút gọn.
d) \(A\in Z\Rightarrow\frac{x-1}{3}\in Z\Rightarrow x=3k+1\)(\(k\in Z\))
3 tháng 3 2022
a: \(B=\left(\dfrac{x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{10}{5\left(x+2\right)}+\dfrac{1}{x-2}\right):\dfrac{x^2-4+6-x^2}{x-2}\)
\(=\left(\dfrac{1}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x+2}+\dfrac{1}{x-2}\right):\dfrac{2}{x-2}\)
\(=\dfrac{1-2x+4+x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2}=\dfrac{-x+7}{2\left(x+2\right)}\)
b: Ta có: |x|=1/2
=>x=1/2 hoặc x=-1/2
Thay x=1/2 vào B, ta được:
\(B=\dfrac{-\dfrac{1}{2}+7}{2\left(\dfrac{1}{2}+2\right)}=\dfrac{13}{10}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{\dfrac{1}{2}+7}{2\left(-\dfrac{1}{2}+2\right)}=\dfrac{5}{2}\)