\(+,x< -2\Rightarrow\left\{{}\begin{matrix}x+2< 0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=-2-x\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow1-3x=5\Rightarrow x=-\frac{4}{3}\left(\text{loại}\right)\)

\(+,x\ge\frac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3\ge0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|x+2\right|=x+2\end{matrix}\right.\Rightarrow3x-1=5\Rightarrow x=2\left(\text{thoa man}\right)\)

\(+,-2\le x< \frac{3}{2}\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=x+2\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow5-x=0\Rightarrow x=0\left(\text{thoa man}\right)\)

\(2.\text{ Ta co:}\left\{{}\begin{matrix}\left|x-102\right|\ge102-x\\\left|2-x\right|\ge x-2\end{matrix}\right.\Rightarrow A\ge102-x+x-2=100.\Rightarrow A_{min}=100.\text{dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}102-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow2\le x\le102\)

Dung mà cx dùng cái này cơ.Tao Bống nè!!!

102 x 2 + 102 x 4 + 102 x 7 +102 x 8 

= 120 x ( 2 x 4 x 7 x 8 ) 

= 120 x 448

= 53760

102x(2+4+7+8)=102x21=2142

102 x 2 + 102 x 4 + 102 x 7 + 102 x 8

= 102 x ( 2 + 4 + 7 + 8 ) 

= 102 x 21

= 2142 

\(A=\dfrac{\left|x-150\right|+102}{\left|x-150\right|+100}=\dfrac{\left|x-150\right|+100+2}{\left|x-150\right|+100}=1+\dfrac{2}{\left|x-150\right|+100}\)

Có: |x - 150| ≥ 0

=>  |x - 150| + 100 ≥ 100

\(\Rightarrow\dfrac{2}{\left|x-150\right|+100}\le\dfrac{1}{50}\)

 \(\Rightarrow A\le\dfrac{51}{50}\)

Dấu = xảy ra khi x = 150

Vậy:...

\(A=\dfrac{\left|x-150\right|+102}{\left|x-150\right|+100}\\ A=\dfrac{\left|x-150\right|+100+2}{\left|x-150\right|+100}\\ A=1+\dfrac{2}{\left|x-150\right|+100}\)

\(\left|x-150\right|\ge0\forall x\\ \Rightarrow\left|x-150\right|+100\ge100\forall x\\ \Rightarrow\dfrac{2}{\left|x-150\right|+100}\le\dfrac{1}{50}\forall x\\ \Rightarrow A=1+\dfrac{2}{\left|x-150\right|+100}\le\dfrac{51}{50}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left|x-150\right|=0\\ \Leftrightarrow x-150=0\\ \Leftrightarrow x=150\)

Vậy GTLN của \(A=\dfrac{51}{50}\) khi x = 150

a) |5/3 - x| - |-5/6| = |-5/9|

=> |5/3 - x| - 5/6 = 5/9

=> |5/3 - x| = 5/9 + 5/6

=> |5/3 - x| = 25/18

=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)

a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)

\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)

b, \(\left|x+\frac{1}{102}\right|+\left|x+\frac{2}{102}\right|+...+\left|x+\frac{100}{102}\right|\ge0\)

\(\Rightarrow102x\ge0\Leftrightarrow x\ge0\)=> Ta có thể phá dấu GTTĐ

\(\Rightarrow x+\frac{1}{102}+x+\frac{2}{102}+...+x+\frac{100}{102}=102x\)

\(\Rightarrow100x+\frac{1+2+3+...+100}{102}=102x\Rightarrow2x\Rightarrow x.\)

\(I=16\times102+24\times98+48\times17\)

\(I=1632+2352+816\)

\(I=4800\)

I = 16 x 102 + 24 x 98 + 48 x 17

I=1632 +2352+816

I=3984+816

I=4800

`16 xx 102 + 24 xx 98 + 48 xx 17`

`= 16 xx 3 xx 34 + 24 xx 2 xx 49 + 48 xx 17`

`= 48 xx 34 + 48 xx 49 + 48 xx 17`

`= 48 xx (34 + 49 + 17)`

`= 48 xx 100`

`= 4800`

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Từ \(\frac{x-2016}{100}\rightarrow\frac{x-2}{2114}\) có tất cả \(1008\) số.

Ta có: \(\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}=1008\)

\(\Leftrightarrow\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}-1008=1008-1008\)

\(\Leftrightarrow\left(\frac{x-2016}{100}-1\right)+\left(\frac{x-2014}{102}-1\right)+...+\left(\frac{x-2}{2114}-1\right)=0\) (*)

\(\Leftrightarrow\frac{x-2116}{100}+\frac{x-2116}{102}+...+\frac{x-2116}{2114}=0\)

\(\Leftrightarrow\left(x-2116\right)\left(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\right)=0\)

Vì \(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\ne0\)

Nên từ pt (*) \(\Leftrightarrow x-2116=0\Leftrightarrow x=2116\)

Vậy...

\(A=\left|x-102\right|+\left|2-x\right|\)

Áp dụng bất đẳng thức:

\(\left|A+B\right|\le\left|A\right|+\left|B\right|\)

\(\Rightarrow A=\left|x-102\right|+\left|2-x\right|\le\left|x-102+2-x\right|\)

\(A\le\left|-100\right|\Rightarrow A\le100\)

Dấu "=" xảy ra khi:

\(-2\le x\le102\)