=2(x^2-5/2x+3/2)

=2(x^2-2*x*5/4+25/16-1/16)

=2(x-5/4)^2-1/8>=-1/8

Dấu = xảy ra khi x=5/4

1: Ta có: \(x^2-2x-5\)

\(=x^2-2x+1-6\)

\(=\left(x-1\right)^2-6\ge-6\forall x\)

Dấu '=' xảy ra khi x=1

2: ta có: \(3x^2+5x-2\)

\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{2}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)

\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{12}\ge-\dfrac{49}{12}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{6}\)

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)

Tìm giá trị nhỏ nhất của biểu thức:

a) Ta có: 

\(M=2x^2+4x+7\)

\(M=2\cdot\left(x^2+2x+\dfrac{7}{2}\right)\)

\(M=2\cdot\left(x^2+2x+1+\dfrac{5}{2}\right)\)

\(M=2\cdot\left[\left(x+1\right)^2+2,5\right]\)

\(M=2\left(x+1\right)^2+5\)

Mà: \(2\left(x+1\right)^2\ge0\forall x\) nên:

\(M=2\left(x+1\right)^2+5\ge5\forall x\)

Dấu "=" xảy ra:

\(2\left(x+1\right)^2+5=5\Leftrightarrow2\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy: \(M_{min}=5\) khi \(x=-1\)

b) Ta có:

\(N=x^2-x+1\)

\(N=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\) nên \(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=" xảy ra: 

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(N_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

Tìm giá trị lớn nhất của biểu thức

a) Ta có: 

\(E=-4x^2+x-1\)

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

\(E=-\left[\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right]\)

\(E=-\left[\left(2x-\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\)

Mà: \(\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\ge\dfrac{15}{16}\forall x\) nên 

\(\Rightarrow E=-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\le-\dfrac{15}{16}\forall x\)

Dấu "=" xảy ra:

\(-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]=-\dfrac{15}{16}\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2-\dfrac{15}{16}=-\dfrac{15}{16}\)

\(\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2=0\Leftrightarrow2x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

Vậy: \(E_{max}=-\dfrac{15}{16}\) khi \(x=\dfrac{1}{16}\)

b) Ta có:

\(F=5x-3x^2+6\)

\(F=-3x^2+5x-6\)

\(F=-\left(3x^2-5x-6\right)\)

\(F=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(F=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{97}{36}\right]\)

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\)

Mà: \(-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\) nên:

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\le\dfrac{97}{36}\forall x\)

Dấu "=" xảy ra:

\(-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}=\dfrac{97}{36}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)

Vậy: \(F_{max}=\dfrac{97}{36}\) khi \(x=\dfrac{5}{6}\)

\(M=2x^2+4x+7\)

\(=2\left(x^2+2x+\dfrac{7}{2}\right)\)

\(=2\left(x^2+2x+1+\dfrac{5}{2}\right)\)

\(=2\left[\left(x+1\right)^2+\dfrac{5}{2}\right]\)

\(=2\left(x+1\right)^2+5\)

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

\(\Rightarrow2\left(x+1\right)^2+5\ge5\forall x\)

\(\Rightarrow M_{min}=5\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Tương tự: \(N=x^2-x+1\)

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

\(\Rightarrow N_{min}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

\(E=-4x^2+x-1\)

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

\(=-4\left[x^2-2.x.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2-\left(\dfrac{1}{8}\right)^2+\dfrac{1}{4}\right]\)

\(=-4\left[\left(x-\dfrac{1}{8}\right)^2+\dfrac{15}{64}\right]\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\)

Vì \(-4\left(x-\dfrac{1}{8}\right)^2\le0\forall x\)

\(\Rightarrow-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\forall x\)

\(\Rightarrow E_{max}=-\dfrac{15}{16}\Leftrightarrow-4\left(x-\dfrac{1}{8}\right)^2=0\Leftrightarrow x=\dfrac{1}{8}\)

Tương tự: \(F=5x-3x^2+6\)

\(=-3x^2+5x+6\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

\(\Rightarrow F_{max}=\dfrac{97}{12}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{5}{6}\)

cau A thay = bằng cộng ạ

A = x² - 6x + 11 

Nhập phương trình vào máy tính lặp 3 lần  dấu =

GTNN của A = 3

B = 2x² + 10x - 1

Nhập phương trình vào máy tính lặp 3 lần dấu =

GTNN của B = \(-\frac{5}{2}\)

C = 5x - x² 

=> C = -x² + 5x

Nhập phương trình vào máy tính lặp 3 lần dấu =

GTLN của C = \(\frac{5}{2}\)

Trả lời

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hok tốt

\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+4\\ A=\left(x-y\right)^2+\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\Leftrightarrow x=y=1\)

\(A=x^4+2x^2-8x+2019\) \(=x^4-2x^2+1+4x^2-8x+4+2014\)

\(=\left(x^2-1\right)^2+4\left(x-1\right)^2+2014\ge2014\forall x\)  

" = " \(\Leftrightarrow x=1\)