GTLN của Q = -1996/1997 <=> x = 0

GTLN của P = -1996/1997 <=> x = 0

k cho mk nha

\(\frac{\left|x\right|+2015}{2016}\) . Có: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2015\ge2015\Rightarrow\frac{\left|x\right|+2015}{2016}\ge\frac{2015}{2016}\)

Dấu = xảy ra khi \(x+2015=0\Rightarrow x=0\)

Vậy \(Min\frac{\left|x\right|+2015}{2016}=\frac{2015}{2016}\) tại \(x=0\)

\(\frac{\left|x\right|+1996}{-1997}\) có \(\left|x\right|\ge0\Rightarrow\left|x\right|+1996\ge1996\Rightarrow\frac{\left|x\right|+1996}{-1997}\le-\frac{1996}{1997}\)

Dấu = xảy ra khi \(\left|x\right|+1996=1996\Rightarrow x=0\) 

Vậy \(Max\frac{\left|x\right|+1996}{-1997}=\frac{1996}{-1997}\) tại \(x=0\)

\(\frac{1996}{\left|x\right|+1997}\) có \(\left|x\right|+1997\ge0\Rightarrow\left|x\right|+1997\ge1997\Rightarrow\frac{1996}{\left|x\right|+1997}\le\frac{1996}{1997}\)

Dấu = xảy ra khi \(x+1997=1997\Rightarrow x=0\)

Vậy \(Max\frac{1996}{\left|x\right|+1997}=\frac{1996}{1997}\) tại \(x=0\)

x=-2000           

ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)

        \(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)

     \(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)  (1)

                     Xét     \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)

                                        từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )

                                                          \(\Rightarrow x=2000\)

GTLN là -1

a) \(\dfrac{1997x1996-1}{1995x1997+1996}=\dfrac{1997x\left(1995+1\right)-1}{1995x1997+1996}\)

\(=\dfrac{1997x1995+1997-1}{1995x1997+1996}=\dfrac{1997x1995+1996}{1995x1997+1996}=1\)

b) \(\dfrac{1997x1996-995}{1995x1997+1002}=\dfrac{1997x\left(1995+1\right)-995}{1995x1997+1002}\)

\(=\dfrac{1997x1995+1997-995}{1995x1997+1002}=\dfrac{1997x1995+1002}{1995x1997+1002}=1\)

KQ = 995

Mình mới giải xong, mất tiêu bài rồi KQ=1000

cộng 1 vào mỗi tỉ số,ta đc:

(x+5)/1995+1+(x+4)/1996+1+(x+3)/1997+1=(x+1995)/5+1+(x+1996)/4+1+(x+1997|/3+1

=>\(\frac{x+5+1995}{1995}+\frac{x+4+1996}{1996}+\frac{x+3+1997}{1997}=\frac{x+1995+5}{5}+\frac{x+1996+4}{4}+\frac{x+1997+3}{3}\)

\(\Rightarrow\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}-\frac{x+2000}{5}-\frac{x+2000}{4}-\frac{x-2000}{3}=0\)

\(\Rightarrow\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

mà bt trong ngoặc thứ 2 khác 0

=>x+2000=0

=>x=-2000