\(A=\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}\)

\(=\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}\)

\(=\frac{x^3+y^3+y^3}{xyz}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)-3xyz}{xyz}\)( HĐT x³ + y³ + z³ - 3xyz chắc biết rồi ha )

\(=\frac{-3xyz}{xyz}=-3\)( do x+y+z = 0 )

à nhầm dấu tí nhé ._.

\(A=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz}{xyz}=3\)

A = x² - 5x + 7 

= ( x² - 5x + 25/4 ) + 3/4

= ( x - 5/2 )² + 3/4 ≥ 3/4 ∀ x 

Dấu "=" xảy ra khi x = 5/2

=> MinA = 3/4, đạt được khi x = 5/2

a, Điều kiện xác định của A là : \(\hept{\begin{cases}3x-6\ne0\\x^2-2x\ne0\end{cases}\Rightarrow\hept{\begin{cases}3\left(x-2\right)\ne0\\x\left(x-2\right)\ne0\end{cases}\Rightarrow x\ne}0;2}\)

Vậy \(x\ne0;2\)

b, \(A=\frac{x}{3x-6}+\frac{2x+3}{x^2-2x}\)

\(=\frac{x}{3\left(x-2\right)}+\frac{2x+3}{x\left(x-2\right)}=\frac{x^2}{3x\left(x-2\right)}+\frac{6x+9}{3x\left(x-2\right)}=\frac{x^2+6x+9}{3x\left(x-2\right)}\)

c, Ta có : A = 0 hay 

\(\frac{x^2+6x+9}{3x\left(x-2\right)}=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)

Vậy x = -3 thì A = 0 

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

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

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

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