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\(3-\frac{x}{5}-x=\frac{x}{x-1}\)
\(\Rightarrow\frac{15\left(x-1\right)}{5\left(x-1\right)}-\frac{x\left(x-1\right)}{5\left(x-1\right)}-\frac{5x\left(x-1\right)}{5\left(x-1\right)}=\frac{5x}{5\left(x-1\right)}\)
\(\Rightarrow15\left(x+1\right)-x\left(x-1\right)-5x\left(x-1\right)=5x\)
\(\Rightarrow15x+15-x^2+x-5x^2+5x=5x\)
Bạn tự làm tiếp theo ha
\(\frac{3-x}{5-x}=\frac{x}{x+1}\)
\(\left(3-x\right)\left(x+1\right)=\left(5-x\right)x\)
\(3\left(x+1\right)-x\left(x+1\right)=5x-x^2\)
\(3x+3-x^2-x=5x-x^2\)
\(2x+3-x^2=5x-x^2\)
\(2x+3=5x\)
\(3=5x-2x\)
\(3x=3\)
\(x=1\)
Vậy x = 1
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/ x - 2 / >= x - 2
/ x - 5 / = / 5 - x / >= 5 - x
=> / x - 5 / + / x - 2 / >= 5 - x + x - 2
/ x - 5 / + / x - 2 / >= 3
Dấu = xảy ra khi :
x - 2 >= 0 => x >= 2
5 - x >= 0 => x <= 3
Vậy x = 2 ; 3
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ta có : x:\(\dfrac{x}{y}\)=\(\dfrac{1}{3}\)
->x.\(\dfrac{y}{x}\)=\(\dfrac{1}{3}\)
->y=\(\dfrac{1}{3}\)
->x-\(\dfrac{3}{\dfrac{1}{3}}\)=\(\dfrac{1}{2}\)
->x = \(\dfrac{19}{2}\)
Vậy......
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Ta có: \(\hept{\begin{cases}8\left(x-2009\right)^2\ge0\Leftrightarrow25-y^2\ge0\\8\left(x-2009\right)^2⋮8\Leftrightarrow25-y^2⋮8\end{cases}}\)
Mà: \(25-y^2\le25\) nên: \(\hept{\begin{cases}0\le25-y^2\le25\\25-y^2⋮8\end{cases}}\)
Ta dễ dàng tìm được: \(25-y^2\in\left\{0;8;16;24\right\}\)
\(\Rightarrow y^2\in\left\{25;17;9;1\right\}\Leftrightarrow y\in\left\{5;\sqrt{17};3;1\right\}\left(y\in N\right)\)
Nên ta chọn: \(5;3;1\).Thay vào tìm được x tương ứng
\(\left(x-1\right)^2+\left|x^2-1\right|=0.\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|x^2-1\right|\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\x^2-1=0\end{cases}\Rightarrow x=1}\)