A.

(1) around

(2) watch

(3) listen

(4)

(5) series

(6)show

(7) station

(8) receive

(9) cities

(10) possible

TH1:a+b+c=0

\(\)\(\Rightarrow\left\{\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)

\(\Rightarrow A=\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}=-1\)

TH2:\(a+b+c\ne0\)

Áp dụng tc dãy tỉ số bằng nhau ta có:

A=\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy A=-1 hoặc A=\(\frac{1}{2}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{91.93}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{91}-\frac{1}{93}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{93}\right)\)

\(=\frac{1}{2}.\frac{92}{93}\)

\(=\frac{46}{93}\)

Vì \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (Tổng 3 góc của tam giác)

\(\Rightarrow\widehat{B}+\widehat{C}=130^0\)

Theo bài ra ta có:\(\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}\)

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{B}+\widehat{C}}{2+3}=\frac{130}{5}=26\)

\(\Rightarrow\left\{\begin{matrix}\widehat{B}=52^0\\\widehat{C}=78^0\end{matrix}\right.\)

Vậy...

Để P nguyên \(\Rightarrow x-2⋮x+1\Rightarrow\left(x+1\right)-3⋮x+1\)

\(\Rightarrow3⋮x+1\)

\(\Rightarrow x+1\inƯ\) của 3

\(\Rightarrow x+1\in\left\{1;3;-1;-3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;-4\right\}\)

Vậy...

a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+..+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)\(\Rightarrow3A=2^{101}-2\)

\(\Rightarrow A=\frac{2^{101}-2}{3}\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:

\(A=\left|2x-2\right|+\left|2x-2013\right|=\left|2x-2\right|+\left|2013-2x\right|\ge\left|2x-2+2013-2x\right|=2011\)Vậy Min A=2011

Dấu "=" xảy ra khi \(\left(2x-2\right)\left(2013-x\right)\ge0\Rightarrow1\le x\le\frac{2013}{2}\)

Vậy Min A=2011\(\Leftrightarrow1\le x\le\frac{2013}{5}\)