a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)

d) \(9x^2=1\)

\(\Leftrightarrow x^2=1:9\)

\(\Leftrightarrow x^2=\dfrac{1}{9}\)

\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Ta có \(x=\dfrac{2016}{x\times\left(x+1\right)\times\left(x+2\right)\times........\times\left(x+2016\right)}\)

\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times...........\times2016}\)

Vì x > 0=> \(\left(x+1\right)\times\left(x+2\right)\times...\times\left(x+2016\right)>1\times2\times...\times2016\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{1}{1\times2\times..........\times2016}\)\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{2016}{1\times2\times......\times2016}\)

\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)

Ta có \(x=\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times....\times\left(x+2016\right)}\)

\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times.....\times2016}\)

Vì x>0=>(x+1)×(x+2)×.............×(x+2016) >\(1\times2\times.....\times2016\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{1}{1\times2\times......\times2016}\)

\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{2016}{1\times2\times......\times2016}\)

\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)

đáp số là 996 đúng ko

dung do bn nguyễn như Quỳnh

\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)

\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)

\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)

\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)

\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)

\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)

\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)

a)Ta thấy:

\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)

\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)

\(=\dfrac{a}{x\left(x+a\right)}\)

\(\Rightarrowđpcm\)

b)Ta thấy:

\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)

\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)

\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)

c)Ta thấy:

\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)

a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)

Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)

\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)

Lời giải:

a) Hiển nhiên vế trái $\geq 0$ do tính chất của trị tuyệt đối.

$\Rightarrow 4x\geq 0\Rightarrow x\geq 0$. Đến đây ta có thể phá bỏ dấu trị tuyệt đối

$|x+\frac{11}{17}|+|x+\frac{2}{17}|+|x+\frac{4}{17}|=4x$

$x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x$

$3x+1=4x$

$x=1$

b) Hiển nhiên vế trái $\geq 0$ nên $11x\geq 0\Rightarrow x\geq 0$

Khi đó:

$|x+\frac{1}{2}|+|x+\frac{1}{6}|+|x+\frac{1}{12}|+...+|x+\frac{1}{110}|=x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}$

$=10x+(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110})$

$=10x+(1-\frac{1}{11})=10x+\frac{10}{11}=11x$

$\Rightarrow x=\frac{10}{11}$

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