ai cao thu ve` toan giai cho minh` voi

T

thaonguyenkmhd

thanks mình nha!!!

A=|x-a|+|x-b|+|x-c|+|x-d|
với a<b<c<d
Tìm min A

A=|x-a|+|x-b|+|x-c|+|x-d|

Ta có
(*)|x-a|+|x-d| = |x-a|+|d-x| \geq |x - a + d - x| = |d-a| = d-a do a<d

dấu "=" xảy ra khi ( x-a )( d-x ) \geq 0

\Leftrightarrow ( x-a )( x-d ) \leq 0

\Leftrightarrow [TEX]\left{\begin{x-a \geq 0}\\{x-d \leq 0} [/TEX] \Leftrightarrow [TEX]\left{\begin{x \geq a}\\{x \leq d} [/TEX] \Leftrightarrow a \leq x \leq d

(*)|x-b|+|x-c| = |x-b|+|c-x| \geq |x - b + c - x| = |c-b| = c-b do b<c

dấu "=" xảy ra khi ( x-b )( c-x ) \geq 0

\Leftrightarrow ( x-b )( x-c ) \leq 0

\Leftrightarrow [TEX]\left{\begin{x-b \geq 0}\\{x-c \leq 0} [/TEX] \Leftrightarrow [TEX]\left{\begin{x \geq b}\\{x \leq c} [/TEX] \Leftrightarrow b \leq x \leq c

\Rightarrow A \geq d-a+c-b = d+c-b-a

\Rightarrow min A = d+c-b-a khi [TEX]\left{\begin{a \leq x \leq d}\\{b \leq x \leq c} [/TEX] \Leftrightarrow b \leq x \leq c
 
T

tranquockhaj

A=|x-a|+|x-b|+|x-c|+|x-d|

Ta có
|x-a|+|x-d| = |x-a|+|d-x| |x - a + d - x| = |d-a| = d-a do a<d

dấu "=" xảy ra khi ( x-a )( d-x ) 0

( x-a )( x-d ) 0

a x d

|x-b|+|x-c| = |x-b|+|c-x| |x - b + c - x| = |c-b| = c-b do b<c

dấu "=" xảy ra khi ( x-b )( c-x ) 0

( x-b )( x-c ) 0

b x c

A d-a+c-b = d+c-b-a

min A = d+c-b-a khi b x c
 
Top Bottom