\(P=2x^2+y^2-10x-2xy+2019\)

\(P=x^2-2xy+y^2+x^2-10x+25+1994\)

\(P=\left(x^2-2xy+y^2\right)+\left(x^2-2\cdot x\cdot5+5^2\right)+1994\)

\(P=\left(x-y\right)^2+\left(x-5\right)^2+1994\ge1994\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=5\end{cases}\Rightarrow}x=y=5}\)

Vậy.....

BĐT Nesbitt  nhé ko phải Nesbit đâu .V
Bđt đấy đây: Cho a,b,c dương

CMR: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)

Giải

Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

      \(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

       \(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

       \(=\frac{1}{2}.\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

Áp dụng bđt Cô-si cho 3 số dương được

\(\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

            \(\ge\frac{1}{2}.3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3.\sqrt[3]{\frac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}-3\)

                 \(=\frac{1}{2}.9.\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}-3\)

                  \(=\frac{9}{2}-3\)

                   \(=\frac{3}{2}\)

Dấu "='' xảy ra <=> a=b=c

Vậy ...........

BĐT Nesbit: Với a,b,c dương:

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

\(BĐT\Leftrightarrow\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\ge\frac{9}{2}\)

\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge9\)

\(\Leftrightarrow2\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge9\)

Dùng bất đẳng thức cô si hai lần vào vế trái sẽ có điều cần chứng minh.

\(\frac{x^4+x^3-x^2-2x-2}{x^4+2x^3-x^2-4x-2}=\frac{\left(x^4-x^2-2\right)+\left(x^3-2x\right)}{\left(x^4-x^2-2\right)+\left(2x^3-4x\right)}\)

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

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

\(F\left(x\right)=\frac{x^4+x^3-x^2-2x-2}{x^4+2x^3-x^2-4x-2}\)

\(=\frac{\left(x^4+x^3+x^2\right)-2x^2-2x-2}{\left(x^4+2x^3+x^2\right)-\left(2x^2+4x+2\right)}\)

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

1+2+3+4+5+...........+10000

=10001.10000:2=10001.5000=50005000

Xong rồi đó

( 1 + 10000 ) x ( 10000 : 2 ) = 50005000 

trừ cho nhau là xong

Nói nghe có vẻ dễ ha Trần Hữu Ngọc Minh 

\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)

\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)

\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)

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

đến đây thì dễ rồi :)

mk không giải được phần sau

ko hiểu

Sao đang x ; y ; z -> a ; b ; c ?

a,b,c la x,y,z danh nham