\(\frac{x^2}{9}+\frac{1}{x^2}=\frac{5}{3}\cdot\left(\frac{x}{3}-\frac{1}{x}\right)\)
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23 tháng 4 2016
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
23 tháng 4 2016
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
19 tháng 9 2019
( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
11 tháng 9 2018
a)
( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0
\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
P/s: đợi xíu làm câu b
11 tháng 9 2018
b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{-1}{x+3}=\frac{1}{2015}\)
\(\Leftrightarrow x+3=-2015\)
\(\Leftrightarrow x=-2018\)
Vậy,.........
31 tháng 3 2017
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
\(\frac{x^2}{9}+\frac{1}{x^2}=\frac{5}{3}\left(\frac{x}{3}-\frac{1}{x}\right)\)
đặt \(\left(\frac{x}{3}-\frac{1}{x}\right)=t\Rightarrow t^2=\left(\frac{x^2}{9}-\frac{2}{3}+\frac{1}{x^2}\right)\Rightarrow\frac{x^2}{9}+\frac{1}{x^2}=t^2+\frac{2}{3}\)
\(\Leftrightarrow t^2-\frac{5}{3}t+\frac{2}{3}=0\Leftrightarrow t^2-2.\frac{5}{6}t+\left(\frac{5}{6}\right)^2=\frac{25}{36}-\frac{24}{36}=\frac{1}{36}=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\left[\begin{matrix}t-\frac{5}{6}=\frac{1}{6}\\t-\frac{5}{6}=-\frac{1}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}t=1\\t=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\frac{x}{3}-\frac{1}{x}=1\\\frac{x}{3}-\frac{1}{x}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\ne0\\x^2-x-3=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne0\\\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\\\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\end{matrix}\right.\)
Kết luận: \(\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\\x=-1\\x=3\end{matrix}\right.\)