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\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a) \(5-\frac{2x}{3}=4x-\frac{1}{-5}\)
\(\frac{75-10x}{15}=\frac{60x+3}{15}\)
75 - 10x = 60x +3
72 = 70x
\(\frac{72}{70}\) = x
x =\(\frac{36}{35}\)
Vậy x = \(\frac{36}{35}\)
b) \(2x-\frac{10}{6}=\frac{-27}{5}-x\)
\(2x-\frac{5}{3}=\frac{-27}{5}-x\)
\(\frac{30x-25}{15}=\frac{-81-15}{15}\)
30x =-96+25
30x =-71
x= -71/30
Vậy x= -71/30
c) \(13x-\frac{2}{2x}+5=\frac{76}{17}\)
13x - 1/x +5 = 76/17
\(\frac{221x-17+85}{17x}=\frac{76x}{17x}\)
221x +68 = 76x
221x-76x =-68
145x =-68
x =\(\frac{-68}{145}\)
Vậy .........
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x^3-8x\left(x+2\right)=6\\ \Leftrightarrow\left(x^2+3x+2\right).\left(x+3\right)-x^3-8x^2-16x=6\\ \Leftrightarrow x^3+6x^2+11x+6-x^3-8x^2-16x-6=0\\ \Leftrightarrow-2x^2-5x=0\\ \Leftrightarrow x.\left(-2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
\(A=\frac{x-2}{x+2}=\frac{x^2-4x+4}{x^2-4}=\frac{x^2-4-4x+8}{x^2-4}=1+\frac{-4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=1-\frac{4}{x+2}\)
Để \(A\in Z\) thì \(\frac{4}{x+2}\in Z\Leftrightarrow x+2\inƯ\left(4\right)\)
\(\Rightarrow x\in\left\{-6;-4;-3;-1;0;2\right\}\)
\(B=\frac{3x-6}{x+6}=\frac{3x+18-24}{x+6}=\frac{3\left(x+6\right)}{x+6}-\frac{24}{x+6}=3-\frac{24}{x+6}\)
Để \(B\in Z\) thì \(\frac{24}{x+6}\in Z\Leftrightarrow x+6\inƯ\left(24\right)\)
\(\Rightarrow x\in\left\{-30;-18;-14;-12;-10;-9;-8;-7;-5;-4;-3;-2;0;2;6;18\right\}\)
\(C=\frac{10-5x}{x-5}=\frac{-\left(5x-25+15\right)}{x-5}=\frac{-5\left(x-5\right)}{x-5}-\frac{15}{x-5}=-5-\frac{15}{x-5}\)
Để \(C\in Z\) thì \(\frac{15}{x-5}\in Z\Leftrightarrow x-5\inƯ\left(15\right)\)
\(\Rightarrow x\in\left\{-10;0;4;6;10;20\right\}\)
\(D=\frac{8x-2}{2-4x}=\frac{-\left(4-8x\right)+2}{2\left(1-2x\right)}=\frac{-4\left(1-2x\right)}{2\left(1-2x\right)}+\frac{2}{2\left(1-2x\right)}=-2+\frac{1}{1-2x}\)
Để \(D\in Z\) thì \(\frac{1}{1-2x}\in Z\Leftrightarrow1-2x\inƯ\left(1\right)\)
\(\Rightarrow x=0\)