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

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

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

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

=> \(1-\frac{x}{3}=\frac{2}{3}\)

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

=> x = 1

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

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

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

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

=> \(x:\frac{1}{2}=\frac{3}{4}\)

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

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

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

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

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

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

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

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

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

\(x=1\)

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

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

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

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

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

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

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

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

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

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

\(\Leftrightarrow\left(x-4\right)^2=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Rightarrow x^2-3x-5x+15+1=0\)

\(\Rightarrow x^2-8x+16=0\)

\(\Rightarrow x^2-2x.4+4^2=0\)

\(\Rightarrow\left(x-4\right)^2=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

\(\Leftrightarrow\left(x-4\right)^2=0\)

\(\Leftrightarrow x=4\)

\(∘backwin\)

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

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

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

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

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

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

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

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

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

\(B<1\)

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)
                                                                      < 
                                                                     

a)(x - 2)(y + 3)= 5
Vì x,y là các số nguyên => x-2,y+3 là các ước nguyên của 5
Ta có bảng sau:
 

b) (x + 1)(y - 5) =-7
Vì x,y là các số nguyên => x+1,y-5 là các ước nguyên của -7
Ta có bảng sau:
 

Chúc bạn học tốt!