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

a)  \(0,75\times26+49\times\frac{3}{4}+\frac{3}{4}\times24+\) \(0,75\)

    \(=\) \(0,75\times\left(26+1\right)+\frac{3}{4}\times\left(49+24\right)\)

     \(=\)  \(\frac{3}{4}\times27+\frac{3}{4}\times73\)

      \(=\) \(\frac{3}{4}\times\left(27+73\right)\)

        \(=\)\(\frac{3}{4}\times100=75\)

b) câu b,c làm tương tự

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 = \(\varnothing\)

Bài 3:

a: \(\frac13+\frac23:x=-7\)

=>\(\frac23:x=-7-\frac13=-\frac{22}{3}\)

=>\(x=\frac23:\frac{-22}{3}=\frac{-2}{22}=-\frac{1}{11}\)

b: \(3\frac12-\frac12x=\frac23\)

=>\(\frac72-\frac{x}{2}=\frac13\)

=>\(7-x=\frac23\)

=>\(x=7-\frac23=\frac{21}{3}-\frac23=\frac{19}{3}\)

c: \(\left\lbrack\left(x+\frac13\right)\cdot\frac34+5\right\rbrack:2=3\)

=>\(\left(x+\frac13\right)\cdot\frac34+5=3\cdot2=6\)

=>\(\left(x+\frac13\right)\cdot\frac34=1\)

=>\(x+\frac13=1:\frac34=\frac43\)

=>\(x=\frac43-\frac13=\frac33=1\)

d: \(\left(1\frac23-x\right)\cdot0,75-2=\frac12\)

=>\(\left(\frac53-x\right)\cdot\frac34=2+\frac12=\frac52\)

=>\(\frac53-x=\frac52:\frac34=\frac52\cdot\frac43=\frac{5\cdot2}{3}=\frac{10}{3}\)

=>\(x=\frac53-\frac{10}{3}=-\frac53\)

e: \(x+75\%\cdot x=-1,6\)

=>\(x+0,75x=-1,6\)

=>1,75x=-1,6

=>\(x=-\frac{160}{175}=-\frac{32}{35}\)

f: \(\left(x-\frac23\right)\left(2x+1\right)=0\)

=>\(\left[\begin{array}{l}x-\frac23=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-\frac12\end{array}\right.\)

g: \(1-\left(2x+\frac12\right)^2=\frac34\)

=>\(\left(2x+\frac12\right)^2=1-\frac34=\frac14\)

=>\(\left[\begin{array}{l}2x+\frac12=\frac12\\ 2x+\frac12=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac12-\frac12=0\\ 2x=-\frac12-\frac12=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\end{array}\right.\)

h: \(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2020}{2021}\)

=>\(1-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2020}{2021}\)

=>\(1-\frac{1}{5x+6}=\frac{2020}{2021}\)

=>\(\frac{1}{5x+6}=\frac{1}{2021}\)

=>5x+6=2021

=>5x=2015

=>x=403