mình ko biết nhưng các bạn k mình nha mình đang âm

k mình nha

Vì | x + 1 |>0 hoặc =0

Suy ra : 5 - | x + 1 | < hoặc = 5

Suy ra : A < hoặc = 0

Suy ra : Amax = 5 khi & chỉ khi x + 1 = 0

suy ra : x = -1

Vậy Amax = 5 khi & chỉ khi x = -1

\(M=x^2+2x+2=\left(x^2+x+x+1\right)+1\)

\(M=x\left(x+1\right)+1\left(x+1\right)+1=\left(x+1\right)\left(x+1\right)+1\)

\(M=\left(x+1\right)^2+1\)

Vì \(\left(x+1\right)^2\ge0\) với mọi x

=>\(\left(x+1\right)^2+1\ge1\) với mọi x

=>GTNN của M là 1

Dấu "=" xảy ra <=> x+1=0<=>x=-1

M_min=1 khi x=-1

a haha đoán bừa mà cũng đúng

a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)

b/

1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Suy ra Min A = 7 <=> x = 2

2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Suy ra Min B = 1/4 <=> x = 1/2

3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)

\(\ge-\frac{9}{2}\)

Suy ra Min N = -9/2 <=> x = 1/2

1) \(A=\left(2x^2+1\right)^4-3\ge0-3=-3\) (do \(\left(2x^2+1\right)^4\ge0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow\left(2x^2+1\right)=0\Leftrightarrow2x^2=-1\Leftrightarrow x^2=-\frac{1}{2}\) (vô lí)

Vậy đề sai ~v  (hay là tui làm sai ta)

1b) \(B=3\left|1-2x\right|-5\ge0-5=-5\)  (do \(\left|1-2x\right|\ge0\forall x\))

Dấu "=" xảy ra khi \(\left|1-2x\right|=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

Vậy \(B_{min}=-5\Leftrightarrow x=\frac{1}{2}\)

@tth T làm cũng ra như vậy @@

\(A=x^2-2x+4\)

\(A=x^2-2x+1+3=\left(x-1\right)^2+3\ge3\)

dấu = xảy ra khi x-1=0

=> x=1

Vậy MinA=3 khi x=1

3: 

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

B3:\(\Rightarrow90.10^n-10^n.10^2+10^n.10-20\Rightarrow10^n.\left(90-10^2\right)+10^n.10-20\)

\(\Rightarrow10^n.\left(90-100\right)+10^n.10-20\Rightarrow-10.10^n+10^n.10-20\Rightarrow-20\)

\(A=-\left(x^2-x+5\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{19}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right]\)

\(=-\left(x-\frac{1}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)

Vậy \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

B2: \(\Rightarrow16x^2-8x-\left(16x^2-8x+1\right)-13\Rightarrow16x^2-8x-16x^2+8x-1-13\Rightarrow-14\)