Question

[bui hang trang]

Bài 1 Tìm x 

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

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

Answer 1

[ST]

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 = \(\frac{22}{45}:2=\frac{11}{45}\)

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

Answer 2

[Bùi Thế Hào]

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