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

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

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

23-20(11-x)=3

=>20(11-x)=20

=>11-x=1

=>x=11-1=10

 công thức diện tích toàn phần của hình lăng trụ đứng 

\(A=2^2+2^4+...+2^{20}\)

\(=2^2\left(1+2^2+...+2^{18}\right)=4\left(1+2^2+...+2^{18}\right)⋮4\)

\(A=2^2+2^4+...+2^{18}+2^{20}\)

\(=2^2\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)

\(=5\left(2^2+2^6+...+2^{18}\right)⋮5\)

\(A=1+3+3^2+...+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=40+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

\(=40\left(1+3^4+3^8\right)⋮40\)

Để ý thấy rằng \(1+3+3^2+3^3=40\)

\(A=1+3+3^2+3^3+...+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

\(=40+3^4\times40+3^8\times40\)

\(=40\left(1+3^4+3^8\right)\)

Do đó A chia hết cho 40

Áp dụng BĐT trị tuyệt đối, ta có:

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

Dấu "=" xảy ra khi: \(\left(x-9\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-9\ge0\\2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-9\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge9\\x\le2\end{matrix}\right.\left(\text{vô lí}\right)\\\left\{{}\begin{matrix}x\le9\\x\ge2\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow2\le x\le9\)

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

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

Nên \(x=0\) là nghiệm phương trình đã cho.