Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)
Lập bảng xét dấu:
TH : Xét x < -2
Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)
-2x = 2\(\dfrac{1}{4}\)
=> x = \(-1\dfrac{1}{8}\) ( loại )
TH 2: \(-2\le x< \dfrac{1}{2}\)
Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)
=> \(2,5=\dfrac{3}{4}\) ( loại )
TH3 : \(x\ge\dfrac{1}{2}\)
x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
2x + 1,5 = \(\dfrac{3}{4}\)
x = -0,375( loại )
vậy ....
b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)
c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)
TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)
\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)
TH2 : \(x-1< 0\Rightarrow x< 1\)
=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)
Vậy x = 1