TÌm số nguyên x biết ( có các bước giải )
a \(\dfrac{x}{2}\) = \(\dfrac{2}{x}\) ; b \(\dfrac{x}{-5}\) = \(\dfrac{-5}{x}\)
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a: \(\Leftrightarrow\left(x+1\right)^2=3^2=9\)
=>x+1=3 hoặc x+1=-3
=>x=2 hoặc x=-4
b: \(\Leftrightarrow\left(x-1\right)^2=16\)
=>x-1=4 hoặc x-1=-4
=>x=5 hoặc x=-3
a: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
x - 1 | 1 | -1 | 3 | -3 |
y + 1 | 3 | -3 | 1 | -1 |
x | 2 | 0 | 4 | -2 |
y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{x-2\left(x+2\right)+1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{-6}{x^2-4}\right):\left(\dfrac{6}{x+2}\right)\)
\(\Rightarrow A=-\dfrac{6}{x^2-4}.\dfrac{x+2}{6}=-\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)6}=-\dfrac{1}{x-2}\)
để A<0 thì :
\(\left\{{}\begin{matrix}x-2\ne0\\x-2\notin Z-\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{3;4;5;6;7;8;9;....n\right\}\)
( Z- là tập hợp số nguyên âm )
Để A có giá trị nguyên thì :
\(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
a) \(\dfrac{x}{2}=\dfrac{2}{x}\)
⇔ \(x^2=4\)
⇒ \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
b) \(\dfrac{x}{-5}=\dfrac{-5}{x}\)
⇔ \(x^2=25\)
⇒ \(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(a,\Rightarrow x^2=2^2\\ \Rightarrow x=2\\ b,x^2=\left(-5\right)^2\\ \Rightarrow x=-5\)