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Bài 1:
\(x=\dfrac{1}{2}\); \(y\) là số nguyên âm lớn nhất nên \(y=-1\). Thay x và y vào A ta được:
\(\dfrac{\dfrac{1}{2}^3-3.\dfrac{1}{2}^2+0,5.\dfrac{1}{2}-\left(-1\right)^2-4}{\dfrac{1}{2}^2+\left(-1\right)}=\dfrac{43}{6}\)
Bài 2: Tìm x
\(\dfrac{x-1}{2004}+\dfrac{x-2}{2003}-\dfrac{x-3}{2002}=\dfrac{x-4}{2001}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2004}+1\right)+\left(\dfrac{x-2}{2003}+1\right)-\left(\dfrac{x-3}{2002}+1\right)=\left(\dfrac{x-4}{2001}+1\right)\)
\(\Leftrightarrow\dfrac{x-2005}{2004}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2002}-\dfrac{x-2005}{2001}=0\)
\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x-2005=0\)
\(\Leftrightarrow x=2005\)
Vậy x=2005

a: \(=\dfrac{x^5}{3x}+\dfrac{12x^2}{3x}+\dfrac{5x}{3x}=\dfrac{1}{3}x^4+4x+\dfrac{5}{3}\)
b: \(=\dfrac{-5x^4}{-5x}-\dfrac{15x^3}{5x}+\dfrac{18x}{5x}=x^3-3x^2+\dfrac{18}{5}\)
c: \(=\dfrac{-x^6}{0.5x}+\dfrac{5x^4}{0.5x}-\dfrac{2x^3}{0.5x}=-2x^5+10x^3-4x^2\)

b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....

b/ M (-4;-2)
Thế xM = -4 vào y = 0,5. x
y = 0,5 . (-4)
y = -2 = y-2
Vậy M (-4; -2) thuộc đồ thị hàm số y = f (x) = 0,5.x


a, \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\)
\(\Rightarrow x=11\)
b, \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{15}\)
\(\Rightarrow x=1\)
c, \(-2^3+0,5x=1,5\)
\(\Rightarrow-8+\dfrac{1}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{2}+8\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{19}{2}\)
\(\Rightarrow x=19\)
1) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\) 2)\(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\) \(\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{11}{12}\) \(\dfrac{19}{15}x=\dfrac{2}{3}-\left(\dfrac{-3}{5}\right)\) => \(x=11\) \(\dfrac{19}{15}x=\dfrac{19}{15}\) => \(x=1\) 3) -23 + 0,5x = 1,5 -8 + 0,5x = 1,5 0,5x = 1,5 - (-8) 0,5x = 9,5 x = 9,5 : 0,5 x = 19
\(\left|0.5x+3\right|+2=3x-4\)
\(\Leftrightarrow\left|0.5x+3\right|=3x-6\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-6-0.5x-3\right)\left(3x-6+0.5x+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2.5x-9\right)\left(3.5x-3\right)=0\end{matrix}\right.\Leftrightarrow x=3.6\)