\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

Dấu \("="\Leftrightarrow x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

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

\(C=4\left(x^2-2x+1\right)-4=4\left(x-1\right)^2-4\ge-4\)

Dấu \("="\Leftrightarrow x=1\)

\(D=\dfrac{1}{-\left(x^2+2x+1\right)+6}=\dfrac{1}{-\left(x+1\right)^2+6}\ge\dfrac{1}{6}\)

Dấu \("="\Leftrightarrow x=-1\)

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

\(A_{min}=-7\) khi \(x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{3}{2}\)

\(C=4\left(x^2-2x+1\right)-4=4\left(x-1\right)^2-4\ge-4\)

\(C_{min}=-4\) khi \(x=1\)

Biểu thức D không tồn tại cả max lẫn min

1.

$A=2x^2-8x+1=2(x^2-4x+4)-7=2(x-2)^2-7$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A\geq 2.0-7=-7$

Vậy $A_{\min}=-7$ khi $x-2=0\Leftrightarrow x=2$

2.

$B=x^2+3x+2=(x^2+3x+1,5^2)-0,25=(x+1,5)^2-0,25\geq 0-0,25=-0,25$

Vậy $B_{\min}=-0,25$ khi $x=-1,5$

3.

$C=4x^2-8x=(4x^2-8x+4)-4=(2x-2)^2-4\geq 0-4=-4$

Vậy $C_{\min}=-4$ khi $2x-2=0\Leftrightarrow x=1$

4. Để $D_{\min}$ thì $5-x^2-2x$ là số thực âm lớn nhất

Mà không tồn tại số thực âm lớn nhất nên không tồn tại $x$ để $D_{\min}$

A\(=2x^2-8x+1\)

=2x(x-4)+1≥1

Min A=1 ⇔x=4

B=\(x^2+3x+2\)

\(=\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{1}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)≥\(-\dfrac{1}{4}\)

Min B=-1/4⇔x=-3/2

C=\(4x^2-8x\)

=\(\left(\left(2x\right)^2-2x.4+16\right)-16\)

=(2x-4)^2 -16≥-16

Min C=-16 ⇔x=2

D=\(\dfrac{1}{-\left(x^2-2x+1\right)+6}\)

=\(\dfrac{1}{-\left(x-1\right)^2+6}\)≥\(\dfrac{1}{6}\)

Min D=1/6 ⇔x=1

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

\(A=2x^2+4x+1=2\left(x^2+2x+1\right)-1=2\left(x+1\right)^2-1\ge-1\)

\(A_{min}=-1\) khi \(x=-1\)

Câu B chỉ có max, ko có min

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

\(B_{max}=\dfrac{25}{4}\) khi \(x=\dfrac{3}{2}\)

Câu C cũng chỉ có max, không có min

\(C=-4x^2+8x=-4\left(x^2-2x+1\right)+4=-4\left(x-1\right)^2+4\le4\)

\(C_{max}=4\) khi \(x=1\)

Câu D cũng chỉ có max, không có min

\(D=\dfrac{3}{4x^2-4x+1+4}=\dfrac{3}{\left(2x-1\right)^2+4}\le\dfrac{3}{4}\)

\(C_{max}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

(4 câu có 3 câu sai đề)

\(A=\dfrac{1}{-x^2+2x-2}\)

A min \(\Leftrightarrow\dfrac{1}{A}\)max

ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)

\(\dfrac{1}{A}\)max= -1 tại x=1

=> A min = -1 tại x=1

\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x² nha bn)

B min \(\Leftrightarrow\dfrac{1}{B}\) max

ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)

\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2

\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2

\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)

A max \(\Leftrightarrow\dfrac{1}{A}\) min

\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)

\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)

\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)

B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)

B max \(\Leftrightarrow\dfrac{1}{B}\) min

\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)

\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)

=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)

Đây chỉ là gợi ý !! bn pải tự lí luận nha

tik