\(A=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}\)

Có: \(3x^2+9x+7=3\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{4}=3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)

Vì: \(3\left(x+\frac{3}{2}\right)^2\ge0,\forall x\)

=> \(3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

=>\(\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}}\le40\)

=> \(1+\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{41}{4}}\le41\)

Vậy GTLN của A là \(\frac{81}{41}\) khi \(x=-\frac{3}{2}\)

HELP ME !!!

B=-3(x²-3x)

B=-3(x²-2\(\frac{3}{2}\)x+\(\frac{9}{4}\)-\(\frac{9}{4}\))

B=-3(x-\(\frac{3}{2}\))²+\(\frac{27}{4}\)

Vậy GTLN của B là \(\frac{27}{4}\)hay 6, 25

Àh, 6,75 nhá, t viết nhầm
 

-6,75 hay -\(\frac{27}{4}\)

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

Tìm GTNN

A = x² - 10x + 3 = ( x² - 10x + 25 ) - 22 = ( x - 5 )² - 22 ≥ -22 ∀ x

Dấu "=" xảy ra khi x = 5

=> MinA = -22 <=> x = 5

B = 3x² + 7x - 2 = 3( x² + 7/3x + 49/36 ) - 73/12 = 3( x + 7/6 )² - 73/12 ≥ -73/12 ∀ x

Dấu "=" xảy ra khi x = -7/6

=> MinB = -73/12 <=> x = -7/6

Tìm GTLN

A = -9x² + 12x - 5 = -9( x² - 4/3x + 4/9 ) - 1 = -9( x - 2/3 )² - 1 ≤ -1 ∀ x

Dấu "=" xảy ra khi x = 2/3

=> MaxA = -1 <=> x = 2/3

B = -2x² - 3x + 7 = -2( x² + 3/2x + 9/16 ) + 65/8 = -2( x + 3/4 )² + 65/8 ≤ 65/8 ∀ x

Dấu "=" xảy ra khi x = -3/4

=> MaxB = 65/8 <=> x = -3/4

\(A=x^2-x+3=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}+3=\left(x-2\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\left(\left(x-2\right)^2\ge0\right)\)

\(\Rightarrow Min\left(A\right)=\dfrac{11}{4}\)

\(B=x^2-4x+1=x^2-4x+4-4+1=\left(x-2\right)^2-3\ge-3\left(\left(x-2\right)^2\ge0\right)\)

\(\Rightarrow Min\left(B\right)=-3\)

Câu C bạn xem lại đề

\(D=3-4x-x^2=3+4-4-4x-x^2=7-\left(x^2+4x+4\right)=7-\left(x+2\right)^2\le7\left(-\left(x+2\right)^2\le0\right)\)

\(\Rightarrow Max\left(D\right)=7\)

\(A=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\in R\)

Vậy GTNN của A là 11/4 khi x=1/2

\(B=x^2-4x+1=\left(x^2-2.x.2+4\right)-3\\ =\left(x-2\right)^2-3\ge\left(-3\right)\forall x\in R\\ Vậy:GTNN.của.B.là\left(-3\right).khi.x=2\)