Vì \(\left(2x+\frac{1}{4}\right)^4\ge0;\left|y+\frac{11}{3}\right|\ge0\)

         Suy ra:\(\left(2x+\frac{1}{4}\right)^4\ge0;\left|y+\frac{11}{3}\right|-1\ge-1\)

Vậy dấu = xảy ra khi \(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{4}=0\\y+\frac{11}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{8}\\y=-\frac{11}{3}\end{cases}}\)

                    Min A=-1 khi \(x=-\frac{1}{8};y=-\frac{11}{3}\)

Giups mik vs

Đáp án D

y = 2 x + 1 1 − x y ' = 3 ( 1 − x ) 2 > 0 ∀ x ∈ [ 2 ; 3 ] ⇒ min [ 2 ; 3 ] y = y ( 2 ) = − 5

Chọn đáp án D

Ta có

nên hàm số  y = 2 x + 1 1 - x luôn đồng biến trên đoạn [2;3]

Suy ra  m i n [ 2 ; 3 ] y = y 2 = - 5

Chọn đáp án D

\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)