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.

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

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

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

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

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

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

Gia sử AB < AC

Kẻ BM,CN // DE , trung tuyến AF

Tam giác BMF = tam giác CNF ( g.c.g)

=> MF = NF

=> AB/AD = AM/AG ; AC/AE = AN/AG

=> AB/AD = AM+AN/AG = AF-MF+AF+NF/AG = 2AF/AG = 3 ( VÌ AF = 3/2.AG )

=> ĐPCM

Tk mk nha

\(a,10n^2+n-10⋮n-1\)

\(10n^2-10n+11n-11+1⋮n-1\)

Do \(10n\left(n-1\right)⋮n-1;11\left(n-1\right)⋮n-1\Rightarrow1⋮n-1\)

\(\Rightarrow n-1\in\left(1;-1\right)\)

\(\Rightarrow n\in\left(2;0\right)\)

\(b,25n^2-97n+11⋮n-4\)

\(25n^2-100n+3n-12+23⋮n-4\)

\(25n\left(n-4\right)+3\left(n-4\right)+23⋮n-4\)

\(\Rightarrow23⋮n-4\)

\(\Rightarrow n-4\in\left(1;-1;23;-23\right)\)

\(\Rightarrow n\in\left(5;3;27;-19\right)\)

1+1=2                457-21=436          545+455=1000

1 + 1 = 2

457 - 21 = 436

545 + 455 = 1000

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

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

Để \(N=\frac{3}{x+1}\)là số nguyên 

\(\Rightarrow3⋮x+1\Rightarrow x+1\in U\left(3\right)=\left\{\pm1;\pm3\right\}\)

bn tự lập bảng ha ~