Biểu thức này không có min và cũng không có max

bạn xem lại đề bài 1 là GTNN hay GTLN nha

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

\(maxN=-\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)

\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

a: Ta có: \(N=-x^2-x-1\)

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

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

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

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

b: ta có: \(B=3x^2+4x-13\)

\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)

\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)

\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{2}{3}\)

bạn ghi đề rõ ràng mình giải cho

cho p/s A=6x+1 / 12x^2+1 tim GTNN va GTLN cua A

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(A=-2x^2+6x-12\)

\(=-2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{15}{2}\)

\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{15}{2}\le-\dfrac{15}{2}\)

\(maxA=-\dfrac{15}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Ta có: \(A=-2x^2+6x-12\)

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

\(=-2\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{15}{4}\right)\)

\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{15}{2}\le-\dfrac{15}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

\(A\ge1\forall x\)

Dấu '=' xảy ra khi x=0

\(B\ge-5\forall x\)

Dấu '=' xảy ra khi x=0

\(A=x^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(A_{min}=1\Leftrightarrow x=0\)

\(B=3x^4-5\ge-5\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(B_{min}=-5\Leftrightarrow x=0\)

=xA=x2+1≥1

Dấu "=" xảy ra ⇔x=0min=1⇔x=0$4-5\ge -$5

Vậy B

A = x² - 8x + 1 = (x² - 8x + 16) - 15 = (x - 4)² - 15

Ta có: (x - 4)² \(\ge\)0 \(\forall\)x

=> (x - 4)² - 15 \(\ge\)-15 \(\forall\) x 

Dấu "=" xảy ra khi: x - 4 = 0 <=> x = 4

vậy Min của A = -15 tại x = 4

B = 9x² - 12x - 2 = 9(x² - 4/3x + 4/9) - 6 = 9(x - 2/3)² - 6

Ta có: (x - 2/3)² \(\ge\)0 \(\forall\)x ---> 9(x - 2/3)² \(\ge\)0 \(\forall\)x

=> 9(x - 2/3)² - 6 \(\ge\)-6 \(\forall\)x

Dấu "=" xảy ra khi: x - 2/3 = 0 <=> x = 2/3

vậy Min của B = -6 tại x = 2/3