Ta có:

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

Áp dụng:

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{625}\right)\)

\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{25^2}\right)\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{24.26}{25^2}\)

\(=\dfrac{1.2.3...24}{2.3.4...25}.\dfrac{3.4.5...26}{2.3.4...25}=\dfrac{1}{25}.\dfrac{26}{2}=\dfrac{13}{25}\)

\(=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{575}{576}.\dfrac{624}{625}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{24.26}{25.25}\)

\(=\dfrac{\left(1.2...24\right).\left(3.4...26\right)}{\left(2.3...25\right).\left(2.3...25\right)}\)

\(=\dfrac{1.26}{25.2}=\dfrac{1.2.13}{25.2}=\dfrac{13}{25}\)

\(\left(x-3\right)^2=\left(1-3x\right)^2\)

=>\(\left(3x-1\right)^2-\left(x-3\right)^2=0\)

=>\(\left(3x-1-x+3\right)\left(3x-1+x-3\right)=0\)

=>(2x+2)(4x-4)=0

=>\(\left[{}\begin{matrix}2x+2=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Cách 1: Chỉ ra tính chất đặc trưng 

\(B=\left\{x=2k;k\in N|k\le6\right\}\)

Cách 2: liệt kê

\(B=\left\{0;2;4;6;8;10;12\right\}\)

Cách 1: Liệt kê: 

`B =` {`0;2;4;6;8;10;12`}

Cách 2: Đặc trưng:

`B =` {`x` thuộc `N | x ⋮ 2`` và `x ≤13`}

\(\dfrac{3}{4}:1\dfrac{1}{14}+\dfrac{-2}{3}:\dfrac{4}{15}\\ =\dfrac{3}{4}:\dfrac{15}{14}+\dfrac{-2}{3}:\dfrac{4}{15}\\ =\dfrac{3}{4}\cdot\dfrac{14}{15}+\dfrac{-2}{3}\cdot\dfrac{15}{4}\\ =\dfrac{7}{10}+\dfrac{-5}{2}\\ =\dfrac{7}{10}+\dfrac{-25}{10}\\ =\dfrac{-18}{10}=-\dfrac{9}{5}\)

\(\dfrac{x-1}{1}+\dfrac{x-1}{2}=\dfrac{x}{3}+\dfrac{x}{4}-\dfrac{7}{12}\\ =>x-1+\dfrac{x}{2}-\dfrac{1}{2}=\dfrac{x}{3}+\dfrac{x}{4}-\dfrac{7}{12}\\ =>\left(x+\dfrac{x}{2}\right)+\left(-1-\dfrac{1}{2}\right)=\left(\dfrac{x}{3}+\dfrac{x}{4}\right)-\dfrac{7}{12}\\ =>\dfrac{3}{2}x-\dfrac{3}{2}=\dfrac{7x}{12}-\dfrac{7}{12}\\ =>\dfrac{3}{2}x-\dfrac{7}{12}x=-\dfrac{7}{12}+\dfrac{3}{2}\\ =>\dfrac{11}{12}x=\dfrac{11}{12}=\\ =>x=\dfrac{11}{12}:\dfrac{11}{12}\\ =>x=1\)

Ta có:

\(\left|x-9\right|+\left|2-x\right|\ge\left|x-9+2-x\right|=\left|-7\right|=7\)

Dấu "=" xảy ra:

\(\left(x-9\right)\left(2-x\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-9\ge0\\2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-9\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow2\le x\le9\)

\(\dfrac{4}{15}< \dfrac{x}{30}< \dfrac{1}{3}\)

=>\(\dfrac{8}{30}< \dfrac{x}{30}< \dfrac{10}{30}\)

=>8<x<10

=>x=9

\(\dfrac{4}{15}< \dfrac{x}{30}< \dfrac{1}{3}\\ =>\dfrac{8}{30}< \dfrac{x}{30}< \dfrac{10}{30}\\ =>8< x< 10\)

`y-3y+7y=30`

`=> (1-3+7) y = 30`

`=> 5y = 30`

`=> y = 30 : 5`

`=> y = 6`

Vậy `y=6`

\(y-3y+7y=30\\ \Rightarrow y.\left(1-3+7\right)=30\\ \Rightarrow5y=30\\ \Rightarrow y=30:5\\ \Rightarrow y=6\)
Vậy \(y=6\)

`x^2 + 12x + 36 - 4x^2`

`= x^2 + 2.x . 6 + 6^2 - (2x)^2`

`= (x+6)^2 - (2x)^2`

`= (x+6+2x)(x+6-2x)`

`= (3x + 6)(6-x)`

`= 3(x + 2)(6-x)`

\(x^2+12x+36-4x^2\)

\(=\left(x+6\right)^2-4x^2\)

\(=\left(x+6+2x\right)\left(x+6-2x\right)=\left(-x+6\right)\left(3x+6\right)=3\left(x+2\right)\left(-x+6\right)\)

\(y-3y+7\cdot7=30\)

=>-2y=30-49=-19

=>\(y=\dfrac{19}{2}\)