1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

 0,5 x + 0,6 ( x - 2 ) = 3

0,5 x + 0.6 x - 1,2 = 3

1,1 x = 4,2

x = \(\frac{42}{11}\)

Kết luận: 

b) \(\frac{1}{3}x-0,5x=0,75\)

\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(-\frac{1}{6}x=\frac{3}{4}\)

\(x=-\frac{9}{2}\)

Kết luận: 

c) \(\frac{3}{-2}x-0,5x=75\%\)

-1,5x - 0,5x = 0,75 

-2x = 0,75 

x = -0,375

Kết luận: 

d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

-0,4 x + 0,25 = 0,75 - 0,75 x 

-0,4 x + 0,75 x = 0,75 - 0,25

0,35 x = 0,5 

x = \(\frac{10}{7}\)

Kết luận: 

\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)

\(\frac{11}{10}x-\frac{6}{5}=3\)

\(\frac{11}{10}x=\frac{21}{5}\)

\(x=\frac{42}{11}\)

\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(\frac{-1}{6}x=\frac{3}{4}\)

\(x=\frac{-9}{2}\)

\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)

\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(-2x=\frac{3}{4}\)

\(x=\frac{-3}{8}\)

\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)

\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)

\(\frac{7}{20}x=\frac{1}{2}\)

\(x=\frac{10}{7}\)

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