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 = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)

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

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)

a)   \(-\frac{7}{2}:\left(3-x\right)-0,75=\frac{1}{4}\)

\(\Leftrightarrow\frac{-7}{2\left(3-x\right)}=\frac{1}{4}+0,75=0,25+0,75=1\)

\(\Leftrightarrow2\left(3-x\right)=-7\)

\(\Leftrightarrow3-x=-\frac{7}{2}\)

\(\Leftrightarrow x=3+\frac{7}{2}\)

\(\Leftrightarrow x=\frac{13}{2}\)

b)  \(\left|2x-\frac{4}{3}\right|-1\frac{1}{3}=-\frac{8}{9}\)

\(\Leftrightarrow\left|2x-\frac{4}{3}\right|=\frac{4}{3}-\frac{8}{9}=\frac{4}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{4}{3}=\frac{4}{9}\\2x-\frac{4}{3}=-\frac{4}{9}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{4}{3}+\frac{4}{9}=\frac{16}{9}\\2x=\frac{4}{3}-\frac{4}{9}=\frac{8}{9}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{9}\\x=\frac{4}{9}\end{cases}}\)

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

1/9 - 0,3 . 5/9 + 1/3 + 5

= 1/9 - 3/10 . 5/9 + 1/3 + 5

= 1/9 - 1/6 + 16/3

= -1/18 + 16/3

= 95/18

0,75 : ( \(\frac{9}{2}\)- 2x ) = \(\frac{-9}{12}\)

\(\frac{3}{4}:\left(\frac{9}{2}-2x\right)=\frac{-9}{12}\)

          \(\frac{9}{2}-2x=\frac{3}{4}:\frac{-9}{12}\)

          \(\frac{9}{2}-2x=-1\)

                     \(2x=\frac{9}{2}-\left(-1\right)\)

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

                       \(x=\frac{7}{2}:2\)

                       \(x=\frac{7}{4}\)

                    Vậy x = \(\frac{7}{4}\)

\(\sqrt{0,36}-\left|-0,75\right|:\left(-1\dfrac{3}{4}\right)\\ =0,6-0,75:\dfrac{-7}{4}\\ =\dfrac{3}{5}-\dfrac{3}{4}\cdot\dfrac{-4}{7}\\ =\dfrac{3}{5}-\dfrac{-12}{28}\\ =\dfrac{3}{5}+\dfrac{3}{7}\\ =\dfrac{36}{35}\)

\(1-\left(\dfrac{5}{9}-\dfrac{2}{3}\right)^2:\dfrac{4}{27}\\ =1-\left(\dfrac{5}{9}-\dfrac{6}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\left(-\dfrac{1}{9}\right)^2\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{81}\cdot\dfrac{27}{4}\\ =1-\dfrac{1}{12}\\ =\dfrac{11}{12}\)