\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)

\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)

suy ra A>B

A = \(-\frac{1.3}{2.2}.-\frac{2.4}{3.3}.\cdot\cdot\cdot-\frac{2013.2015}{2014.2014}=-\frac{\left(1.2.3...2013\right).\left(3.4.5....2015\right)}{\left(2.3....2014\right).\left(2.3....2014\right)}=-\frac{2.2015}{2014}=-\frac{4030}{2014}

A<B

mình đúng nha

a, Có : (1/60)^200 = [(1/2)^4]^200 = (1/2)^800

Vì 0 < 1/2 < 1 nên (1/2)^800 > (1/2)^1000

=> (1/16)^200 > (1/2)^1000

Tk mk nha

a) \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}\)

a) \(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}\)

\(\left(\frac{1}{2}\right)^{1000}=\frac{1}{2^{1000}}\)

có : \(16^{200}=\left(2^4\right)^{200}=2^{800}\)

ta thấy \(2^{800}< 2^{1000}\)

\(\Rightarrow16^{200}< 2^{1000}\)

\(\Rightarrow\frac{1}{16^{200}}>\frac{1}{2^{1000}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{100}\)

-A =( 1- 1/2 )(1 -1/3).....(1 -1/10)

    = 1/2 . 2/3 ..... 9/10

    = 1/10

-A = 1/10 nên A = -1/10

Vì 1/10 < 1/9 nên -1/10 > -1/9

Vậy A > -1/9

\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=-\frac{1}{2}.-\frac{2}{3}...-\frac{9}{10}\)

\(=\frac{-\left(1.2...9\right)}{2.3...10}=\frac{-1}{10}\)

A<-1/9

@Nguyễn Việt Lâm

https://hoc24.vn/id/2782086

Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)

Thật vậy, BĐT trên tương đương:

\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)

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

\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)

Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)

\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)

\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)