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\(\dfrac{x}{2}+\dfrac{3x}{5}-\dfrac{6}{5}=3\Leftrightarrow\dfrac{11x}{10}=3+\dfrac{6}{5}=\dfrac{21}{5}\)
\(\Rightarrow11x=42\Leftrightarrow x=\dfrac{42}{11}\)
\(\frac{1}{2}\)x + \(\frac{3}{5}\)( x - 2 ) = 3
\(\frac{1}{2}\)x + \(\frac{3}{5}\) x - \(\frac{3}{5}\). 2 = 3
x \((\)\(\frac{1}{2}\)+ \(\frac{3}{5}\) \()\)- \(\frac{6}{5}\) = 3
x . \(\frac{11}{10}\) - \(\frac{6}{5}\) = 3
x . \(\frac{11}{10}\) = 3 + \(\frac{6}{5}\) = \(\frac{21}{5}\)
x = \(\frac{21}{5}\) : \(\frac{11}{10}\) = \(\frac{42}{11}\)
NẾU CÓ GÌ SAI SÓT MONG BẠN THÔNG CẢM
(-75)+(-7)^2+|-75|+49-(-4)^3
=(-75)+49+75+49+4^3
= -75+49+75+49+64
= (-75+75)+49+49+64
= 0+49+49+64
= 49+49+64
= 2x49+64
= 98+64
= 162
a/ \(3x+2xy=7\)
\(\Leftrightarrow x\left(2y+3\right)=7\)
\(\Leftrightarrow x;2y+3\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\2y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\2y+3=-7\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\2y+3=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\2y+3=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=-\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
b/ \(3x-5xy=11\)
\(\Leftrightarrow x\left(3-5y\right)\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\3-5y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\3-5y=-11\end{matrix}\right.\\\left\{{}\begin{matrix}x=11\\3-5y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-11\\3-5y=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-\dfrac{8}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=\dfrac{14}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=\dfrac{2}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
a) Vì 12 ⋮ 3x + 1 => 3x + 1 ∊ Ư(12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12} => 3x ∊ {-13;-7;-5;-4;-3;-2;0;1;2;3;5;11}. Vì 3x ⋮ 3 => 3x ∊ {-3;0;3} => x ∊ {-1;0;1}. Vậy x ∊ {-1;0;1}. b) 2x + 3 ⋮ 7 => 2x + 3 ∊ B(7) = {...;-21;-14;-7;0;7;14;21;...}. Vì 2x ⋮ 2 mà 3 lẻ nên khi số lẻ trừ đi 3 thì 2x mới ⋮ 2 => 2x + 3 lẻ => 2x + 3 ∊ {...;-35;-21;-7;7;21;35;...} => 2x ∊ {...;-38;-24;-10;4;18;32;...} => x ∊ {...;-19;-12;-5;2;9;16;...} => x ⋮ 7 dư 2 => x = 7k + 2. Vậy x = 7k + 2 (k ∊ Z)
Ta có: \(100^{2013}=100.100....100=\overline{100...}\)(Chữ số đầu là 1, còn lại là 0)
\(\Rightarrow100^{2013}+2=\overline{100...2}\).
Ta thấy \(\overline{100...2}\)có tổng các số hạng là 3. Mà \(3⋮3\)(Hiển nhiên)
\(\Rightarrow\overline{100...2}⋮3\Rightarrow100^{2013}+2⋮3\)(đpcm).
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2020}{2021}\Leftrightarrow\dfrac{x}{x+1}=\dfrac{2020}{2021}\Rightarrow2021x=2020x+2020\Leftrightarrow x=2020\)
3x = 3/4 x - 1,2
3/4x - 3x = 1,2
-2,25x = 1,2
x = 1,2 : (-2,25)
=> x = -8/15