Tìm x \(\in\)Z biết |x-a| = a ( với a \(\in\)Z)
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ĐKXĐ: \(x\ge0;x\ne25\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5+7}{\sqrt{x}-5}=1+\dfrac{7}{\sqrt{x}-5}\)
Để \(A\in\mathbb{Z}\) thì: \(\dfrac{7}{\sqrt{x}-5}\) nhận giá trị nguyên
\(\Rightarrow 7\vdots\sqrt{x}-5\)
\(\Rightarrow\sqrt{x}-5\inƯ\left(7\right)\)
\(\Rightarrow\sqrt{x}-5\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{6;12;4;-2\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{4;6;12\right\}\)
\(\Rightarrow x\in\left\{16;36;144\right\}\left(tm\right)\)
Vậy \(A\in \mathbb{Z}\) khi \(x\in\left\{16;36;144\right\}\)
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\(A\cap B=\left\{{}\begin{matrix}x>m\\x\le\dfrac{2m-1}{3}\end{matrix}\right.\left(1\right)\)
\(TH1:m< \dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}< 0\)
\(\Leftrightarrow\dfrac{m-1}{3}< 0\)
\(\Leftrightarrow m< 1\)
\(\left(1\right)\Leftrightarrow A\cap B=\left\{x\in Z|m< x\le\dfrac{2m-1}{3}\right\}\)
\(TH2:m>\dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}>0\)
\(\Leftrightarrow\dfrac{m-1}{3}>0\)
\(\Leftrightarrow m>1\)
\(\left(1\right)\Leftrightarrow A\cap B=\varnothing\)
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có ; +)x+8=6 +) x+8= - 6
x= 6-8 x= - 6-8
x=6+(-8) x=-6+(-8)
x=-2 X= - 14
có +) x-a=a +) x- a =- a
x=a+a x=-a+a
x=c x=c
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đk x khác 0
\(A=4+\dfrac{6}{\sqrt{x}}\Rightarrow\sqrt{x}\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)
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Ta có : \(A=\dfrac{x^2+2x+1-4x-4+4}{x+1}\)
\(=\dfrac{\left(x+1\right)^2-4\left(x+1\right)+4}{x+1}=x+1-4+\dfrac{4}{x+1}\)
- Để A là số nguyên
\(\Leftrightarrow x+1\inƯ_{\left(4\right)}\) ( Do x là số nguyên )
\(\Leftrightarrow x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;1;-3;3;-5\right\}\)
Vậy ....
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a. |x + 8| = 6
TH1: x + 8 = -6
x = -14
TH2: x + 8 = 6
x = -2
b. 1 < |x - 2| < 4
\(\Rightarrow\left|x-2\right|\in\left\{2;3\right\}\)
TH1: |x - 2| = 2
\(\Leftrightarrow\orbr{\begin{cases}x-2=-2\\x-2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
TH2: |x - 2| = 3
\(\Leftrightarrow\orbr{\begin{cases}x-2=-3\\x-2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
c. |x - a| = a
TH1: x - a = -a
x = 0
TH2: x - a = a
x = 2a
a, =) x+6= -6 hc 6 =) +) x+6=6=)x=0 +) x+6= -6 =)x=-1 mk lm dc mỗi í a thôi à :) chúc pn hok tốt
|x-a|=a
=>x-a=a hay x-a=-a
x=a+a x=-a+a
x=2a x=0
Vậy x=2a hoặc x=0
(+) l x - a l = x - a khi x >= a thay vào ta có:
x - a = a
=> x = 2a
(+) lx - a l = a- x thay vòa ta cso :
a - x = a
-x = 0
x = 0