Ta có: \(N=-9x^2+6x+5\)

\(=-\left(9x^2-6x-5\right)\)

\(=-\left(9x^2-6x+1-6\right)\)

\(=-\left(3x-1\right)^2+6\le6\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

Ta có: F = 5 + 6x + 9x^2

=> F = (3x)^2 + 2.3x.1 + 1^2 + 4

=> F = (3x+1)^2 +4 \(\ge4\). Dấu "=" xảy ra \(\Leftrightarrow3x+1=0\Rightarrow x=\frac{-1}{3}\)

Vậy: GTNN của F = 4 khi x = -1/3

\(F=5+6x+9x^2\)'

\(F=9x^2+6x+1+4\)

\(F=\left(3x+1\right)^2+4\)

\(Do\left(3x+1\right)^2\ge0\Rightarrow F\ge4\)

Dấu "=" xảy ra khi 3x + 1 =0

   <=> 3x = -1

   <=> x = -1/3

Vậy Min F = 4 khi x = -1/3

thanks 2 bãn nhiều

1) \(M=9x^2-6x+6=\left(9x^2-6x+1\right)+5=\left(3x-1\right)^2+5\ge5\)

\(minM=5\Leftrightarrow x=\dfrac{1}{3}\)

2) \(M=5-2x-x^2=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)

\(maxM=6\Leftrightarrow x=-1\)

3) \(N=5+6x-9x^2=-\left(9x^2-6x+1\right)+6=-\left(3x-1\right)^2+6\le6\)

\(maxN=6\Leftrightarrow x=\dfrac{1}{3}\)

Đặt A 9x² + 6x - 1 = 9x² + 6x + 1 - 2 = (3x + 1)² - 2 \(\ge\)-2

=> Min A = -2

Dấu "=" xảy ra <=> 3x + 1 = 0 

<=> x = -1/3

Vậy Min A = -2 <=> x = -1/3 

=-1/3 nha

Trả lời:

1, \(9x^2+6x-1=9x^2+6x+1-2=\left(3x+1\right)^2-2\ge-2\forall x\)

Dấu "=" xảy ra khi 3x + 1 = 0 <=> x = - 1/3

Vậy GTNN của bt = - 2 khi x = - 1/3

a) \(A=\sqrt{4x^2+4x+2}=\sqrt{4x^2+4x+1+1}=\sqrt{\left(2x+1\right)^2+1}\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow A\ge\sqrt{1}=1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)

b) \(B=\sqrt{2x^2-4x+5+1}=\sqrt{2x^2-4x+2+3+1}=\sqrt{2\left(x^2-2x+1\right)+4}\)

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

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow B\ge\sqrt{4}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(minB=2\Leftrightarrow x=1\)

Mơn bạn nha

_hì^^!!

TL:

a,\(-\left(x^2-2x+1\right)+1\)1

\(-\left(x-1\right)^2+1\) \(\le\) 1

=>giá trị lớn nhất của biểu thức là 1

vậy........

b,\(-\left(9x^2+6x+1\right)+20\) 

   \(-\left(3x+1\right)^2+20\) 

  \(\le20\) 

=>giá trị lớn nhất cuar biểu thức là 20

vậy.........

hc tốt

13V&#x1EAD;y&#xA0;MaxB=20&#xA0;khi&#xA0;x=&#x2212;13" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">b) B=19−6x−9x2B=20−1−6x−9x2B=20−(1+6x+9x2)B=20−(1+3x)2Do (1+3x)2≥0∀x⇒B=20−(1+3x)2≤20∀xDấu "=" xảy ra khi:(1+3x)2=0⇔1+3x=0⇔3x=−1⇔x=−13Vậy MaxB=20 khi x=−13