\(\frac{5}{3}-\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{7}{6}\)

=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)

=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{2}\)

=> \(\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{3}:\frac{1}{2}=\frac{1}{3}\cdot2=\frac{2}{3}\)

=> \(1-\frac{x}{3}=\frac{2}{3}\)

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

=> x = 1

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

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

=> \(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}=\frac{9}{4}\)

=> \(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)

=> \(x:\frac{1}{2}=\frac{3}{4}\)

=> \(x=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)

\(\frac{5}{3}-\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{7}{6}\)

\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)

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

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

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

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

\(x\times\frac{1}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{1}{3}\)

\(x=1\)

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

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

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

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

\(x:\frac{1}{2}+\frac{3}{2}=\frac{9}{4}\)

\(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)

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

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

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

\(∘backwin\)

\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)

\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)

\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)

\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(B<1-\)\(\dfrac{1}{2021}\)

\(B<\)\(\dfrac{2020}{2021}\)

\(\dfrac{2020}{2021}< 1\)

\(B<1\)

a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7

b) Ta thấy: 1/2^2 < 1/2.3
                  1/3^2 < 1/3.4
                        ...
                  1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
     Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
         =1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
         =1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
                                                                      < 
                                                                     

mình chỉ có thể giúp bạn hai câu đầu thôi thông cảm nha

( ko có dấu gạch nên ghi là "gạch " ) 

gạch x + 1 gạch = 2^3 - 1

gạch x + 1 gạch = 8 - 1

gạch x + 1 gạch = 7

x + 1 = 7 hoặc x + 1 = - 7 

 x = 7 - 1 hoặc x = ( -7 ) - 1

 x = 6 hoặc x = - 8

Vậy  x = 6 hoặc x = - 8

gạch 2 - x gạch = 5

2 - x = 5 hoặc 2 - x = - 5

x = 2 - 5 hoặc x = 2 -  ( -5 ) 

x = ( -3 ) hoặc x = 7

Vậy x = ( -3 ) hoặc x = 7