\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\)\(\frac{b.\left(a+b\right)}{ab.\left(a+b\right)}+\frac{a.\left(a+b\right)}{ab.\left(a+b\right)}\ge\frac{4ab}{ab.\left(a+b\right)}\)

\(\Leftrightarrow\) \(ab+b^2+a^2+ab\ge4ab\)

\(\Leftrightarrow\)\(a^2+b^2+2ab-4ab\ge0\)

\(\Leftrightarrow\)\(a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2\ge0\) ( đpcm )

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\left(true\right)\)

\(N=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2014}-1\right)\left(\frac{1}{2015}-1\right)\)

\(=-\frac{1}{2}\left(-\frac{2}{3}\right)....\left(-\frac{2013}{2014}\right)\left(-\frac{2014}{2015}\right)\)

\(=\frac{1}{2015}\)

\(N=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)....\left(\frac{1}{2015}-1\right)\)

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

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

\(=\frac{-1}{2015}\)

a) Ta có : 11.13.15.17 + 1137

            = ...5 + ....7

            = ...2

=> 11.13.15.17 + 1137 không là số chính phương

b) Ta có : 25.26.27.28 - 857

            = ...0 - ....7

            = ...3

=>  25.26.27.28 - 857 không là số chính phương 

Ta có:\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng) \(\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)(1)

Mặt khác: \(\frac{\left(x+y\right)^2}{2}\ge2xy\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng) \(\Rightarrow\frac{\left(x+y\right)^2}{2}\ge2xy\) (2)

Từ (1) và (2) => đpcm

A)\(\left|x\right|=\left|\frac{-5}{7}\right|\Rightarrow\left|x\right|=\frac{5}{7}\)

                                     \(\Rightarrow\orbr{\begin{cases}x=\frac{5}{7}\\x=\frac{-5}{7}\end{cases}}\)

B)Mình ko hiểu đề bài cho lắm. Sorry nha!!

\(a,|x|=|-\frac{5}{7}|\)

\(\Leftrightarrow|x|=\frac{5}{7}\)

\(\Leftrightarrow x=\pm\frac{5}{7}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{7}\\x=-\frac{5}{7}\end{cases}}\)

\(b,x=a-\frac{4}{5}\)

\(A,\)Để X là số dương \(\Rightarrow x>0\Rightarrow a-\frac{4}{5}>0\Rightarrow a>\frac{4}{5}\)

 B)Để X là số âm \(\Rightarrow x< 0\Rightarrow a-\frac{4}{5}< 0\Rightarrow a< \frac{4}{5}\)

C)Để X không phải số dương hay số âm \(\Rightarrow x=0\Rightarrow a-\frac{4}{5}=0\Rightarrow a=\frac{4}{5}\)

với a,b,c lớn thì \(\frac{1}{\left(a+2b+c\right)^3}\) nhỏ, \(a^3+8b^3+c^3\) lớn => P ko có max 

\(P=\frac{a^3+8b^3+c^3}{\left(a+2b+c\right)^3}=\left(\frac{a}{a+2b+c}\right)^3+\left(\frac{2b}{a+2b+c}\right)^3+\left(\frac{c}{a+2b+c}\right)^3\)

Đặt \(\left(x;y;z\right)\rightarrow\left(\frac{a}{a+2b+c};\frac{2b}{a+2b+c};\frac{c}{a+2b+c}\right)\)\(\Rightarrow\)\(x+y+z=1\)

\(P=x^3+y^3+z^3=\frac{x^4}{x}+\frac{y^4}{y}+\frac{z^4}{z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\frac{\frac{\left(x+y+z\right)^4}{9}}{x+y+z}=\frac{1}{9}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\) hay \(a=2b=c\)

A= 2xyz - xy - yz - zx +1 
= z(xy-1) - (xy-1) + zy(x-1) - z(x-1) 
= (z-1)(xy-1) + z(x-1)(y-1) 

Do x,y,z >1 nên A>0 suy ra đpcm

nguồn:Cho A= 2xyz - xy - yz - zx +1. Chứng minh A>0 với mọi x>1, y>1, z>1.

A= 2xyz - xy - yz - zx +1 
= z(xy-1) - (xy-1) + zy(x-1) - z(x-1) 
= (z-1)(xy-1) + z(x-1)(y-1) 

Do x,y,z >1 nên A>0 suy ra đpcm