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4 tháng 1 2023
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
18 tháng 11 2017
a, 4C = 12|x|+8/4|x|-5 = 3 + 23/|x|-5 <= 3 + 23/0-5 = -8/5
=> C <= -2/5
Dấu "=" xảy ra <=> x=0
Vậy Min ...
b, Để C thuộc N => 3|x|+2 chia hết cho 4|x|-5
=> 4.(3|x|+2) chia hết cho 4|x|-5
<=> 12|x|+8 chia hết cho 4|x|-5
<=> 3.(|x|+5) + 23 chia hết cho 4|x|-5
=> 23 chia hết chi 4|x|-5 [ vì 3.(4|x|-5) chia hết cho 4|x|-5 ]
Đến đó bạn tìm ước của 23 rùi giải
LM
7 tháng 6 2017
a, \(A=\dfrac{2x^3+x^2+2x+4}{2x+1}\\ =\dfrac{2x^3+x^2+2x+1+3}{2x+1}\\ =\dfrac{\left(2x+1\right)\left(x^2+1\right)+3}{2x+1}\\ =x^2+1+\dfrac{3}{2x+1}\)
Để \(A\in Z\) thì \(2x+1\inƯ\left(3\right)\)= \(\left\{\pm1;\pm3\right\}\)
=> \(2x\in\left\{-4;-2;0;2\right\}\) \(\Rightarrow x\in\left\{-2;-1;0;1\right\}\)
b, Để A vô nghĩa thì 2x+1=0 \(\Leftrightarrow\)x=\(\dfrac{-1}{2}\)
22 tháng 7 2018
\(a.\)
\(P=\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\left(\dfrac{1}{x^2}+\dfrac{x^2}{x^2}\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+\dfrac{x}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2}.\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{x+1}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2\left(x^2+2x+1\right)}+\dfrac{2}{x\left(x+1\right)^2}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2}{x^3+2x^2+x}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x\left(x^3+2x^2+x\right)}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x^4+2x^3+x^2}\right].\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+1+2x}{x^4+2x^3+x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+2x+1}{x^2\left(x^2+2x+1\right)}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{1}{x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x-1}{x^5}\)
NN
8 tháng 12 2017
a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)
\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2}{1-2x}\)
b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)
\(\Rightarrow6=-2+4x\)
\(\Rightarrow4x=6-\left(-2\right)\)
\(\Rightarrow4x=6+2\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8:4\)
\(\Rightarrow x=2\)
Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)
c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)
\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{1;0\right\}\)
Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)
K
8 tháng 12 2017
a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
= \(\left(\dfrac{1}{1-x}+\dfrac{2}{1+x}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{x^2-1}{1-2x}\)
= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
=\(\dfrac{2}{1-2x}\)
b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
=> 2 . 3 = -2 (1 - 2x) (tích chéo)
=> 6 = -2 + 4x
=> 6 + 2 - 4x = 0
=> 8 - 4x = 0
=> 4x = 8
=> x = 2 (thỏa mãn đkxđ)
Vậy để M = \(\dfrac{-2}{3}\) thì x = 2
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
DT
1
28 tháng 9 2016
Ta có:
\(B=\frac{2x^3+x^2+2x+4}{2x+1}=\frac{x^2.\left(2x+1\right)+2x+1+3}{2x+1}\)
\(B=\frac{\left(2x+1\right).\left(x^2+1\right)+3}{2x+1}\)
\(B=\frac{\left(2x+1\right).\left(x^2+1\right)}{2x+1}+\frac{3}{2x+1}\)
\(B=x^2+1+\frac{3}{2x+1}\)
Do x nguyên nên x2 + 1 nguyên
Để B nguyên thì \(\frac{3}{2x+1}\) nguyên
\(\Rightarrow3⋮2x+1\)
\(\Rightarrow2x+1\in\left\{1;-1;3;-3\right\}\)
\(\Rightarrow2x\in\left\{0;-2;2;-4\right\}\)
\(\Rightarrow x\in\left\{0;-1;1;-2\right\}\)
Vậy \(x\in\left\{0;-1;1;-2\right\}\)
\(\dfrac{3}{\sqrt{x}-4}\in Z\Leftrightarrow3⋮\sqrt{x}-4\\ \Leftrightarrow\sqrt{x}-4\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\)
ĐK: \(x\ge0;x\ne16\)
\(\dfrac{3}{\sqrt{x}-4}\in Z\)
\(\Leftrightarrow\sqrt{x}-4\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\inƯ_3=\left\{1;3;5;7\right\}\)
\(\Leftrightarrow x\inƯ_3=\left\{1;9;25;49\right\}\)