Ta có M= |x+2|+|x-9|+|x+1945|

           = |x+1945|+|x+2|+|9-x|

Vì |x+1945|>= x+1945

     |x-2|>= 0

     |9-x|>= 9-x

nên M=|x+1945|+|x-2|+|9-x| >= x+1945+0+9-x =1954

Suy ra min M =1954 (=) x=2

Vậy min M =1954 (=) x=2

\(A=\left|x-2\right|+\left|x-9\right|+\left|1945-x\right|\)

\(x\ge9\)

\(A=\left|1945-x-11\right|\)

\(A=\left|1945-11-11\right|\left(min/x=11\right)\)

\(A=\left|1945\right|\)

GTNN = \(\left|1945\right|\)(:

\(A=\left|x-9\right|+\left|10-x\right|\)

Ta có:

\(\left|x-9\right|\ge x-9\forall x\)

\(\left|10-x\right|\ge10-x\forall x\)

\(\Rightarrow\left|x-9\right|+\left|10-x\right|\ge x-9+10-x\)

\(\Rightarrow A\ge1\)

Dấu bằng xảy ra

\(\Leftrightarrow\orbr{\begin{cases}x-9\ge0\\10-x\ge0\end{cases}\Leftrightarrow9\le x\le10}\)

Vậy minA = 1 \(\Leftrightarrow9\le x\le10\)

\(B=\left|x-1945\right|+\)\(\left|x-1954\right|\)

Ta có:

\(\left|x-1945\right|\ge x-1945\forall x\)

\(\left|x-1954\right|\ge1954-x\forall x\)

\(\Leftrightarrow\left|x-1945\right|+\left|x-1954\right|\ge x-1945+1954-x\)

\(\Leftrightarrow B\ge9\)

Dấu bằng xảy ra

\(\Leftrightarrow\orbr{\begin{cases}x-1945\ge0\\1954-x\ge0\end{cases}\Leftrightarrow1945\le x\le1954}\)

Vậy minB = 9 \(\Leftrightarrow1945\le x\le1954\)

a) \(M=2022-\left|x-9\right|\le2022\)

\(maxM=2022\Leftrightarrow x=9\)

b) \(N=\left|x-2021\right|+2022\ge2022\)

\(minN=2022\Leftrightarrow x=2021\)

hình như bạn cho đề sai

đúng đè mà!

$\textbf{a)}$

$A=(x+1)(x+2)(x+3)(x+4)+12$

$=\big[(x+1)(x+4)\big]\big[(x+2)(x+3)\big]+12$

$=(x^2+5x+4)(x^2+5x+6)+12.$

Đặt $t=x^2+5x+5.$

Khi đó $x^2+5x+4=t-1,\qquad x^2+5x+6=t+1.$

Suy ra $A=(t-1)(t+1)+12$

$\phantom{A}=t^2+11$

$\phantom{A}=(x^2+5x+5)^2+11\ge11.$

Dấu ``='' xảy ra khi $x^2+5x+5=0$

$\Leftrightarrow x=\dfrac{-5\pm\sqrt5}{2}.$

Vậy $\min A=11.$

$\textbf{b)}$

$M=(x+1)^4+(x+3)^4.$

Đặt $t=x+2.$

Khi đó $M=(t-1)^4+(t+1)^4$

$=2t^4+12t^2+2$

$=2(t^2+3)^2-16$

$\ge2\cdot3^2-16$

$=2.$

Dấu ``='' xảy ra khi $t=0$

$\Leftrightarrow x=-2.$

Vậy $\min M=2$, đạt được khi $x=-2.$

ta có 

\(A=\left|x-8\right|+\left|x+2\right|+\left|x+5\right|+\left|x+7\right|\ge\left|-x+8-x-2+x+5+x+7\right|=18\)

Dấu bằng xảy ra khi \(-5\le x\le-2\)

\(B=\left|x+3\right|+\left|x-5\right|+\left|x-2\right|\ge\left|x+3-x+5\right|+\left|x-2\right|=8+\left|x-2\right|\ge8\)

Dấu bằng xảy ra khi \(x=2\)

\(C=\left|x+5\right|-\left|x-2\right|\le\left|x+5+2-x\right|=7\)

Dấu bằng xảy ra khi \(x\ge2\)

Nguyễn Minh Quang sai dấu câu A rồi

a: Khi x=-2 thì \(M=3-\left(-2-1\right)^2=3-9=-6\)

Khi x=0 thì \(M=3-\left(0-1\right)^2=2\)

Khi x=3 thì \(M=3-\left(3-1\right)^2=3-2^2=-1\)

b: Để M=6 thì \(3-\left(x-1\right)^2=6\)

\(\Leftrightarrow\left(x-1\right)^2=-3\)(loại)

c: \(M=-\left(x-1\right)^2+3\le3\forall x\)

Dấu '=' xảy ra khi x=1

a, Thay x=-2 vào M ta có:
\(M=3-\left(-2-1\right)^2=3-\left(-3\right)^2=3-9=-6\)

 Thay x=0 vào M ta có:
\(M=3-\left(0-1\right)^2=3-\left(-1\right)^2=3-1=2\)

 Thay x=3 vào M ta có:
\(M=3-\left(3-1\right)^2=3-2^2=3-4=-1\)

b, Để M=6 thì:

\(3-\left(x-1\right)^2=6\\ \Leftrightarrow\left(x-1\right)^2=-3\left(vô.lí\right)\)

c, Ta có: \(\left(x-1\right)^2\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

\(\Rightarrow M=3-\left(x-1\right)^2\le3\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

Vậy \(M_{max}=3\Leftrightarrow x=1\)