1) `(x-3)^4 >=0`

`2.(x-3)^4>=0`

`2.(x-3)^4-11 >=-11`

`=> A_(min)=-11 <=> x-3=0<=>x=3`

2) `|5-x|>=0`

`-|5-x|<=0`

`-3-|5-x|<=-3`

`=> B_(max)=-3 <=>x=5`.

Bài 1: 

Ta có: \(\left(x-3\right)^4\ge0\forall x\)

\(\Leftrightarrow2\left(x-3\right)^4\ge0\forall x\)

\(\Leftrightarrow2\left(x-3\right)^4-11\ge-11\forall x\)
Dấu '=' xảy ra khi x=3

Bài 2: 

Ta có: \(\left|5-x\right|\ge0\forall x\)

\(\Leftrightarrow-\left|5-x\right|\le0\forall x\)

\(\Leftrightarrow-\left|5-x\right|-3\le-3\forall x\)

Dấu '=' xảy ra khi x=5

\(A=12-\left(2,5-y\right)^4\le12\)

\(maxA=12\Leftrightarrow y=2,5\)

\(B=10-\left(3+4y\right)^2-\left(x-2y\right)^2\le10\)

\(maxB=10\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-\dfrac{3}{4}\end{matrix}\right.\)

1/  196

2/  5/4

3/  1/3

\(A=\dfrac{27-12x}{x^2+9}=\dfrac{x^2-12x+36-\left(x^2+9\right)}{x^2+9}=\dfrac{\left(x-6\right)^2}{x^2+9}-1\ge-1\)

\(A_{min}=-1\Leftrightarrow x=6\)

\(A=\dfrac{27-12x}{x^2+9}=\dfrac{4\left(x^2+9\right)-\left(4x^2+12x+9\right)}{x^2+9}=4-\dfrac{\left(2x+3\right)^2}{x^2+9}\le4\)

\(A_{max}=4\Leftrightarrow x=\dfrac{-3}{2}\)

De P lon nhat thi 540 : (x-6) lon nhat. De 540:(x-6) lon nhat thi x-6 nho nhat. x-6 nho nhat th x-6=1=>x=1+6=7

De P nho nhat thi 540 :(x-6) nho nhat. De 540 nho nhat thi x-6 lon nhat. de x-6 lon nhat thi x-6=540=>x=546

\(A=-\left|x-3.5\right|+0.5\le0.5\forall x\)

Dấu '=' xảy ra khi x=3,5

\(B=-\left|1.4-x\right|-2\le-2\forall x\)

Dấu '=' xảy ra khi x=1,4

ai giúp mk giải bài này mk k cho

a) \(L=\frac{3}{4}-\left|x-\frac{1}{2}\right|\le\frac{3}{4}\forall x\)

GTLN của L = 3/4 khi x = 1/2.

b) 2X + 3 chia hết cho X - 1

=> 2X - 2 + 5 chia hết cho X - 1

=> 2*(X - 1) + 5 chia hết cho X - 1

=> 5 chia hết cho X - 1

=> X - 1 là U(5) = {-5;-1;1;5}

=> X = -4; 0; 2; 6.

Vậy có 4 giá trị của X là : -4; 0; 2; 6 thì 2X + 3 chia hết cho X - 1.

gia tri lon nhat la 6

A=4-x²+3x

=-x²+3x+4

=\(-x^2+3x-\)\(\frac{9}{4}+\frac{25}{4}\)

=\(-\left(x^2-3x+\frac{9}{4}\right)+\frac{25}{4}\)

\(=\frac{25}{4}-\left(x-\frac{3}{2}\right)^2\)

\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le0\) voi moi x

\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le\frac{25}{4}\)

Vay GTLN la : \(\frac{25}{4}\)

Dau "=" xay ra khi : \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)