S = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) +......+ \(\dfrac{4}{2012.2014}\) + \(\dfrac{4}{2014.2016}\)

S = 2. ( \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+.....+\dfrac{2}{2012.2014}+\dfrac{2}{2014.2016}\))

S = 2.( \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}+.....+\dfrac{1}{2014}-\dfrac{1}{2016}\))

S = 2.(\(\dfrac{1}{2}-\dfrac{1}{2016}\))

S = 2. \(\dfrac{1007}{2016}\)

S = \(\dfrac{1007}{1008}\)

\(\dfrac{3}{2}-|x-\dfrac{3}{4}|=\sqrt{\dfrac{9}{16}}\)

 ⇔ \(\dfrac{3}{2}-|x-\dfrac{3}{4}|=\dfrac{3}{4}\)

⇔ \(|x-\dfrac{3}{4}|=-\dfrac{3}{4}\)

⇔ \(x-\dfrac{3}{4}=-\dfrac{3}{4}\) hoặc      \(x-\dfrac{3}{4}=\dfrac{3}{4}\)

⇔ \(x\)          \(=0\)            ⇔  \(x\)          \(=\dfrac{3}{2}\)

Vậy pt có nghiệm : \(x=0\) hoặc \(x=\dfrac{3}{2}\)

\(\left(3x-1\right)^6=\left(3x-1\right)^4\\ =>\left(3x-1\right)^6-\left(3x-1\right)^4=0\\ =>\left(3x-1\right)^4\left[\left(3x-1\right)^2-1\right]=0\\ =>\left[{}\begin{matrix}\left(3x-1\right)^4=0\\\left(3x-1\right)^2-1=0\end{matrix}\right.=>\left[{}\begin{matrix}\left(3x-1\right)=0\\3x-1=1\\3x-1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)

Lời giải:
Gọi tổng trên là $A$
\(A=\frac{1}{\frac{1\times 2}{2}}+\frac{1}{\frac{2\times 3}{2}}+\frac{1}{\frac{3\times 4}{2}}+\frac{1}{\frac{3\times 4}{2}}+...+\frac{1}{\frac{9\times 10}{2}}\)

\(=\frac{2}{1\times 2}+\frac{2}{2\times 3}+\frac{2}{3\times 4}+....+\frac{2}{9\times 10}\)

\(=2\times (\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+...+\frac{10-9}{9\times 10})\)

\(=2\times (1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10})\)

\(=2\times (1-\frac{1}{10})=2\times \frac{9}{10}=\frac{9}{5}\)

5555

DĐeo

|7x-5y|, |2z-3x| ≥ 0

Để |7x-5y| + |2z-3x| = 0

=> 7x - 5y = 2z - 3x = 0

=> 7x = 5y ; 2z = 3x

=> \(\dfrac{x}{5}\)=\(\dfrac{y}{7}\)

      \(\dfrac{z}{3}\)=\(\dfrac{x}{2}\)

=> \(\dfrac{x}{10}\)=\(\dfrac{y}{14}\)=\(\dfrac{z}{15}\)

=> (\(\dfrac{x}{10}\))² = \(\dfrac{xy}{140}\)= \(\dfrac{yz}{210}\)=\(\dfrac{xz}{150}\)=\(\dfrac{xy+yz+xz}{140+150+210}\)= 4

=> x = 20, y = 28, z = 30