25+12+2018=2055

lễ giáng sinh hqua rùi nhé bn

..........................#rrtt

#froggen

Trả lời :

25 + 12 + 2018 = 2055

Merry Christmas

\(A=\left(x-3\right)\left(7-x\right)=-x^2+10x-21=-\left(x^2-10x+25\right)+4\)

\(A=-\left(x-5\right)^2+4\le4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x-5\right)^2=0\)\(\Leftrightarrow\)\(x=5\)( thỏa mãn \(3\le x\le7\) ) 

... 

Còn cách này hay hơn nhé :)) dùng Cosi 

Vì \(3\le x\le7\) nên \(A=\left(x-3\right)\left(7-x\right)\ge0\)

\(\Rightarrow\)\(\sqrt{A}=\sqrt{\left(x-3\right)\left(7-x\right)}\le\frac{x-3+7-x}{2}=\frac{4}{2}=2\)\(\Leftrightarrow\)\(A=2^2=4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x-3=7-x\)\(\Leftrightarrow\)\(x=5\) ( thỏa mãn \(3\le x\le7\) ) 

\(a-b+b-c=\sqrt{2}+1+\sqrt{2}-1\Rightarrow a-c=2\sqrt{2}\)

Vậy \(2A=2a^2+2b^2+2c^2-2ab-2bc-2ca\)

\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)

\(=\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2\)

\(=2+2\sqrt{2}+1+2-2\sqrt{2}+1+8\)

\(=14\)

Do đó: \(A=\frac{14}{2}=7\)

* t sẽ chứng minh đề thiếu điều kiện \(n>0\)

ĐKXĐ : \(n>0\) hoặc \(n< -1\)

+) Nếu \(n>0\) ta có : 

\(\frac{1}{\sqrt{n^2+1}}< \frac{1}{\sqrt{n^2}}=\frac{1}{\left|n\right|}=\frac{1}{n}\)

\(\frac{1}{\sqrt{n^2+2}}< \frac{1}{n}\)

\(\frac{1}{\sqrt{n^2+3}}< \frac{1}{n}\)

\(............\)

\(\frac{1}{\sqrt{n^2+n}}< \frac{1}{n}\)

\(\Rightarrow\)\(P=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\frac{1}{\sqrt{n^2+3}}+...+\frac{1}{\sqrt{n^2+n}}>\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}\)

\(=n.\frac{1}{n}=1\)

\(\Rightarrow\)\(P< 1\)

+) Nếu \(n< -1\) ta có : 

\(\frac{1}{\sqrt{n^2+1}}< \frac{1}{\sqrt{n^2}}=\frac{1}{\left|n\right|}=\frac{1}{-n}\)

\(\frac{1}{\sqrt{n^2+2}}< \frac{1}{-n}\)

\(\frac{1}{\sqrt{n^2+3}}< \frac{1}{-n}\)

\(............\)

\(\frac{1}{\sqrt{n^2+n}}< \frac{1}{-n}\)

\(\Rightarrow\)\(P=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\frac{1}{\sqrt{n^2+3}}+...+\frac{1}{\sqrt{n^2+n}}< \frac{1}{-n}+\frac{1}{-n}+\frac{1}{-n}+...+\frac{1}{-n}\)

\(=n.\frac{1}{-n}=-1\)

\(\Rightarrow\)\(P< -1\)

Vậy nếu \(n>0\) thì \(P< 1\) , nếu \(n< -1\) thì \(P< -1\)

hehe :)) 

tuyệt :v

\(4\left(x-5y\right)+3\left(3x+4\right)=7\Leftrightarrow4x-20y+9x+12=7\Leftrightarrow13x-20y=-5\)(1)

\(5\left(x-3y\right)-2\left(3x-y\right)=3\Leftrightarrow5x-15y-6x+2y=3\Leftrightarrow-x-13y=3\)

\(\Leftrightarrow13x+169y=-39\)(2)

Từ (1) và (2), ta có: \(\hept{\begin{cases}13x-20y=-5\\13x+169y=-39\end{cases}}\Leftrightarrow\hept{\begin{cases}13x+169y-\left(13x-20y\right)=-39-\left(-5\right)\\13x+169y=-39\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}189y=-34\\x+13y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{-34}{189}\\x=\frac{-125}{189}\end{cases}}\)