Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)

Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)

\(=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}\)

\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{x}+\frac{y}{z}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)

\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

a) \(a^2+b^2+1\ge ab+a+b\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra <=> a=b=1.

b) \(a^2-2a+6b+b^2=-10\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\). Mà \(\left(a-1\right)^2\ge0;\left(b+3\right)^2\ge0\forall a;b\)

Nên \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-3\end{cases}}}\). KL: ...

Gọi giao điểm của tia AE và tia CD là F. 

Dễ thấy: Tứ giác ABDC là hình vuông => AB=BD=DC=CA

Xét \(\Delta\)ABE và \(\Delta\)FDE có: ^ABE = ^FDE (=90⁰), BE=DE; ^AEB = ^FED => \(\Delta\)ABE = \(\Delta\)FDE (g.c.g)

=> AB=FD (2 cạnh tương ứng) => FD=CD => D là trung điểm CF.

Xét \(\Delta\) CMF vuông tại M có trung tuyến MD => MD = CD => DM=DC=DB

\(\Rightarrow\widehat{BMC}=\widehat{DMB}+\widehat{DMC}=\frac{180^0-\widehat{BDM}}{2}+\frac{180^0-\widehat{CDM}}{2}=135^0\)

=> ^KMN = 45⁰. Lại có: \(\Delta\)CDM cân tại D có trung tuyến DN => DN vuông góc CM => ^MNK = 90⁰

Suy ra: \(\Delta\)MNK vuông cân tại N => ^MKN = 45⁰.Hay ^BKD = 45⁰.

Vậy ^BKD = 45⁰.  

\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)

\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)

\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)

\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)

\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)

\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)

\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)

a/ Ta có \(A=\frac{\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}}{1-\frac{x}{x+2}}\)với \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)

\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}}{\frac{x+2-x}{x+2}}\)

\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2x-4}{x^2-4}}{\frac{2}{x+2}}\)

\(A=\frac{\frac{x-x-6}{x^2-4}}{\frac{2}{x+2}}\)

\(A=\frac{-6}{x^2-4}.\frac{x+2}{2}\)

\(A=\frac{-3}{x-2}\)

b/ Ta có \(x=-4\)thoả mãn ĐKXĐ

Vậy với \(x=-4\):

\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{1}{2}\)

c/ Khi \(A\inℤ\)

=> \(\frac{-3}{x-2}\inℤ\)

=> \(-3⋮\left(x-2\right)\)

=> x - 2 là ước của -3

Ta có bảng sau:

Mà ĐKXĐ \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)

=> \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)

Vậy khi \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)thì \(A\inℤ\).