`2+4+6+...+300`

`(300+2) . [(300-2):2+1] : 2`

`= 302 . (298 : 2 + 1) : 2`

`= 302 . (149 + 1) : 2`

`= 302 . 150 : 2`

`= 22650``

Số số hạng là \(\dfrac{300-2}{2}+1=\dfrac{298}{2}+1=150\left(số\right)\)

Tổng của dãy số là (300+2)x150:2=302x75=22650

\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\\ \Rightarrow\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\\\Rightarrow\dfrac{7}{4}:x=\dfrac{1}{6} \\ \Rightarrow x=\dfrac{7}{4}:\dfrac{1}{6}\\ \Rightarrow x=\dfrac{21}{2}\)

Vậy \(x=\dfrac{21}{2}\)

\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\)

=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\)

=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{4}{6}\)

=> \(\dfrac{7}{4}:x=\dfrac{1}{6}\)

=> \(x=\dfrac{7}{4}:\dfrac{1}{6}\)

=> x = \(\dfrac{7}{4}.6\)

=> \(x=\dfrac{21}{2}\)

Vậy ...

\(\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}-2\dfrac{1}{4}:1,\left(3\right)\)

\(=\dfrac{3}{64}-\dfrac{9}{4}:\dfrac{4}{3}\)

\(=\dfrac{3}{64}-\dfrac{27}{16}=\dfrac{3}{64}-\dfrac{108}{64}=-\dfrac{105}{64}\)

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1.5+\dfrac{-3}{5}:x\right)=0\left(x\ne0\right)\\ TH1:\dfrac{3}{4}x-\dfrac{9}{16}=0\\ =>\dfrac{3}{4}x=\dfrac{9}{16}\\ =>x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{3}{4}\left(tm\right)\\ TH2:1,5+\dfrac{-3}{5}:x=0\\ =>\dfrac{3}{5}:x=\dfrac{3}{2}\\ =>x=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{2}{5}\left(tm\right)\)

\(x=\dfrac{3}{4}\) và \(x=\dfrac{2}{5}\)

\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)

<=> \(\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\\ TH1:x+\dfrac{5}{3}=0\\ =>x=\dfrac{-5}{3}\\ TH2:x-\dfrac{5}{4}=0\\ =>x=\dfrac{5}{4}\)

Vậy: ...

\(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\)

=>\(\dfrac{5}{6}x=-\dfrac{4}{15}+\dfrac{2}{5}=\dfrac{2}{15}\)

=>\(x=\dfrac{2}{15}:\dfrac{5}{6}=\dfrac{2}{15}\cdot\dfrac{6}{5}=\dfrac{12}{75}=\dfrac{4}{25}\)

\(\left(-8\dfrac{2}{5}\right):\left(-2\dfrac{4}{5}\right)=\dfrac{-42}{5}:\dfrac{-14}{5}=\dfrac{42}{14}=3\)

`(x-5)(x-7) = 0`

`<=> x-5 = 0` hoặc `x - 7 = 0`

`<=> x = 5` hoặc `x = 7`

Vậy ` x = 5` hoặc `x = 7`

(x-5)(x-7)=0

=>\(\left[{}\begin{matrix}x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)