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Ta có:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Leftrightarrow x=305\)
Bài 1:
\(\left(\frac{2}{5}\right)^2+5\frac{1}{2}\left(4,5-2\right)+\frac{2^3}{-4}\)
\(=\frac{4}{25}+\frac{11}{2}\cdot\frac{5}{2}-2\)
\(=\frac{4}{25}+\frac{55}{4}-2\)
\(=\frac{1191}{100}\)
Bài 2:
\(\left(x-0,2\right)^{10}+\left(y+3,10\right)^{20}=0\)
Ta có: (x-0,2)^10 >/ 0
(y+3,10) >/ 0
=> (x-0,2)^10 =0
x- 0,2 =0
x= 0,2
và (y+ 3,10)^20 =0
y+ 3,10 = 0
y = -3,10
Vậy x= 0,2; y= -3,10
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
Ể? \(x^2+x+1=0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\left(VL\right)\) rồi mà SP.
a) \(\left|2x-3\right|-\frac{1}{3}=0\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{10}{3}\\2x=\frac{8}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)
b) \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{7}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}\)
c) \(3-\left|2x+1,5\right|=\frac{5}{4}\)
\(\Leftrightarrow\left|2x+\frac{3}{2}\right|=\frac{7}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{2}=\frac{7}{4}\\2x+\frac{3}{2}=-\frac{7}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{13}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{13}{8}\end{cases}}\)
a. \(\left|2x-3\right|-\frac{1}{3}=0\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)
b. \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{7}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}\)
c. \(3-\left|2x+1,5\right|=\frac{5}{4}\)
\(\Leftrightarrow\left|2x+\frac{3}{2}\right|=\frac{7}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{2}=\frac{7}{4}\\2x+\frac{3}{2}=-\frac{7}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{13}{8}\end{cases}}\)
x=15/8