cau ket ban khong

My home village is now different from the village in my father’s story. in his story, my village is very poor.Everyone  live  a  hard  life that  time.  People  lack  of  food,  of  money.  Now,  it  is different. People don’t live a hard life anymore. They have enough food, enough money, everyone  got  a  house,  and  many  more  things.  The  only  things  that doesn’t  change  is sentiment. Everyone always be very happy. They live together, help each other when they need, that isa luxury.That makes my village a wonderful place for everyone. People are just farmer, but they solidarity in the face of penury. And when they have money in their hands, they still solidarity they are friendly with others, they are helpful, they are good people. That’s about people.Landscape is still the old days. That river, that mountain. Nature is harmony with people. People live with nature, people product things of nature, and nature always open-handed with people, give them resource for themto build their house. That is what people seek, and they live with it very well.That’s  all  I  can  say  about  my  home  village,  in  my  opinion,  my  village  is  a wonderful place. War does not determine who is right, only who is left:” Ask not what your country can do for you but what you can do for your country”.

số đó là

190

ai ủng hộ

mình mình

ủng hộ 

lại cho

So do la:190

K mình nha!!!Mình ủng hộ lại cho!!!!!DATTEBAYO!!!!

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)

\(=\left(x-y+2\right)^2-9\)

\(=\left(x-y+2\right)^2-3^2\)

\(=\left(x-y-1\right)\left(x-y+5\right)\)

nhớ nha

\(x^2-2xy+y^2+4x-4y-5\)

\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)

\(=\left(x-y+2\right)^2-9\)

\(=\left(x-y+2-3\right)\left(x-y+2+3\right)\)

\(=\left(x-y-1\right)\left(x-y+5\right)\)

a) \(x^2+2x+1\)

\(=\left(x+1\right)^2\)

b) \(x^2-6x+9\)

\(=\left(x-3\right)^2\)

c) \(x^2+4x+4\)

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

d) \(x^3+9x^2+27x+27\)

\(=\left(x+3\right)^3\)

a) \(25x^2-2=0\)

\(=>\left(5x\right)^2-\left(\sqrt{2}\right)^2=0\)

\(=>\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)=0\)

\(=>\hept{\begin{cases}5x-\sqrt{2}=0\\5x+\sqrt{2}=0\end{cases}}\)

\(=>\hept{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)

b) \(10x-x^2-25=0\)

\(=>-x^2-5x-5x-25=0\)

\(=>-x\left(x+5\right)-5\left(x+5\right)=0\)

\(=>\left(x+5\right)\left(-x-5\right)=0\)

\(=>\hept{\begin{cases}x+5=0\\-x-5=0\end{cases}}\)

\(=>\hept{\begin{cases}x=-5\\x=-5\end{cases}}\)

r5te45tey45