a, \(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=0+\frac{1}{16}\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

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

\(\Rightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{3}=\frac{-1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{11}{12}\\\frac{1}{x}=\frac{5}{12}\end{cases}\Rightarrow\orbr{\begin{cases}11x=12\\5x=12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}}\)

b, \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

Đặt S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)

2S = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

S = \(\frac{22}{45}:2=\frac{11}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}\Rightarrow x=\frac{23}{11}\)

a/ (1/x -2/3)²=1/16=(1/4)²

Có 2 trường hợp:

+/ 1/x -2/3= - 1/4

<=> 1/x =2/3 -1/4 = 5/12

=> x₁=12/5

+/ 1/x - 2/3 =1/4

<=> 1/x = 2/3 +1/4= 11/12

=> x₂=12/11

b/ Ta có: 

2/(1.2.3)=1/(1.2) - 1/2.3 ;  2/(2.3.4)=1/2.3 -1/3.4 ; ...; 2/(8.9.10)=1/8.9 -1/9.10

=> (1/1.2.3 + 1/2.3.4 +...+1/8.9.10)=23/45

<=> (1/1.2 -1/2.3 +1/2.3 -1/3.4 +...+1/8.9-1/9.10).x/2=23/45

<=> (1/1.2 -1/9.10).x/2 =23/45

<=> x.11/45=23/45

=> x=23/11

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

\(A=\frac{22}{45}:2=\frac{11}{45}\)

thay A vào ta được

\(\frac{11}{45}.x=\frac{23}{45}\)

        \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

x = 23/11

bạn vào câu hỏi tương tự tham khảo nha

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

c) a=1;b=2;c=3;d=4

\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

hay \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

Vậy \(x=\frac{23}{11}\)

mơn a nèk

\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}x=1\)

\(\Rightarrow x=2\)

=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45

=>(1/1.2-1/2.3+1/2.3-1/3.4+....+1/8.9-1/9.10).x=22/45

=>(1/1.2-1/9.10).x=22/45

=>22/45.x=44/45

=>x=2

4+2^2+2^3+....+2^20=2^n

=>2^2+2^2+2^3+....+2^20=2^n

đặt 2^2+2^3+....+2^20

=>2A-A=2^21-2^2

khi đó A=2^2+2^21-2^2=2^21=2^n

=>n=21

2.(4+2^2+2^3+2^4+...+2^20)=2^3+2^3+2^4+2^5+...+2^21=>4+2^2+2^3+...+2^20=2^3+2^4+2^5+...+2^21-(4+2^2+2^3+2^4+...+2^20)=2^21-2^2-4=2097144

Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)

=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)

2B = \(\frac{22}{45}\)

B = \(\frac{22}{45}:2\)

=> B = \(\frac{11}{45}\)

Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)

=> x = \(\frac{22}{45}:\frac{11}{45}\)

=> x = \(\frac{2}{1}\)

Ta có:

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)

\(\frac{11}{45}.x=\frac{23}{45}\)

 \(x=\frac{23}{45}\div\frac{11}{45}\)

\(x=\frac{23}{11}\)

=> \(x=\frac{23}{11}\)