\(\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ac+1}{a}\)

\(\Leftrightarrow a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}\)

\(\Leftrightarrow\hept{\begin{cases}a-c=\frac{1}{c}-\frac{1}{b}\\b-c=\frac{1}{a}-\frac{1}{c}\\c-a=\frac{1}{b}-\frac{1}{a}\end{cases}\Leftrightarrow\hept{\begin{cases}a-c=\frac{b-c}{bc}\left(1\right)\\b-c=\frac{c-a}{ca}\left(2\right)\\c-a=\frac{a-b}{ab}\left(3\right)\end{cases}}}\)

Nhân (1);(2) và (3) theo vế \(\left(a-c\right)\left(b-c\right)\left(c-a\right)=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{a^2b^2c^2}\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(1-\frac{1}{a^2b^2c^2}\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)

\(\Rightarrow a=b\)hoặc \(b=c\)hoặc \(c=a\)

Với \(a=b\)thay vào \(\left(1\right)\)ta đc:\(b=c\Rightarrow a=b=c\)

Với \(b=c\)thay vào \(\left(2\right)\)ta đc\(c=a\Rightarrow a=b=c\)

Với \(c=a\)thay vào\(\left(3\right)\)ta đc \(a=b\Rightarrow a=b=c\)

\(\Rightarrow a=b=c\)

Nguồn:https://olm.vn/hoi-dap/detail/50048198023.html

a) Ta có:

\(A=1+2+2^2+2^3+...+2^{2015}\)

\(A=\left(1+2+2^2\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)-2^{2016}\)

\(A=7+...+7\cdot2^{2014}-2^{2016}\)

\(A=7\cdot\left(1+...+2^{2014}\right)-2^{2016}\)

Lại có: \(2^4\equiv2\left(mod7\right)\Leftrightarrow\left(2^4\right)^{504}=2^{2016}\equiv2\left(mod7\right)\)

\(\Rightarrow A\equiv-2\left(mod7\right)\)

Vậy A chia 7 dư -2 hoặc 5

b) \(PT\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow x\in\left\{0;-2-;-1\right\}\)

=> Tổng các nghiệm là: -3

chữ hơi xấu : v

\(a,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)\)

\(=\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x+y-x+y\right)^2\)

\(=y^2\)

\(b,\left(x+2\right)\left(x^2-2x+4\right)+\left(1-x\right)\left(1+x+x^2\right)\)

\(=\left(x^3+8\right)+\left(1-x^3\right)\)

\(=x^3+8+1-x^3\)

\(=9\)