\(K=\frac{-7}{-2x^2+8x-60}\)

\(K=\frac{-7}{-2\left(x^2-4x+4-26\right)}\)

\(K=\frac{7}{2\left(x-2\right)^2-56}\)

Ta có : \(2\left(x-2\right)^2-56\ge-56\)

\(\Rightarrow K_{max}=\frac{-7}{56}\Leftrightarrow x=2\)

\(L=\frac{8}{-3x^2+9x-40}\)

\(L=\frac{8}{-3\left(x^2-3x+\frac{9}{4}+\frac{133}{12}\right)}\)

\(L=\frac{-8}{3\left(x-\frac{3}{2}\right)^2+\frac{133}{4}}\)

Ta có : \(3\left(x-\frac{3}{2}\right)^2+\frac{133}{4}\ge\frac{133}{4}\)

\(\Rightarrow L_{max}=-\frac{8.4}{133}=-\frac{32}{133}\Leftrightarrow x=\frac{3}{2}\)

a. Ta có : \(A=\frac{8x^2-9}{x^2+3}=\frac{8x^2+24-33}{x^2+3}=8-\frac{33}{x^2+3}\)

Để Amin thì \(\frac{33}{x^2+3}_{max}\) mà \(\frac{33}{x^2+3}\le11\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+3=3\Leftrightarrow x=0\)

Vậy Amin = 8 - 11 = - 3 <=> x = 0

b. Ta có : \(B=\frac{3x^2-6x+40}{x^2-2x+5}=\frac{3\left(x^2-2x+5\right)+25}{x^2-2x+5}=3+\frac{25}{x^2-2x+5}\)

Để Bmax thì \(\frac{25}{x^2-2x+5}=\frac{25}{\left(x-1\right)^2+4}_{max}\)

mà \(\frac{25}{\left(x-1\right)^2+4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2+4=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Bmax \(=3+\frac{25}{4}=\frac{37}{4}\)  <=> x = 1

\(P=\dfrac{2\left(x^2+2\right)+x^2-4x+4}{x^2+2}=2+\dfrac{\left(x-2\right)^2}{x^2+2}\ge2\)

\(P=\dfrac{5\left(x^2+2\right)-2x^2-4x-2}{x^2+2}=5-\dfrac{2\left(x+1\right)^2}{x^2+2}\le5\)

\(M=\frac{2x^2+4x+60}{x^2+2x+4}=\frac{2\left(x^2+2x+4\right)+52}{x^2+2x+4}=2+\frac{52}{x^2+2x+4}=2+\frac{52}{\left(x+1\right)^2+3}\)

Để M đạt GTNN => \(\frac{52}{\left(x+1\right)^2+3}\)đạt GTLN

=> \(\left(x+1\right)^2+3\)(*) đạt GTNN

\(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+3\ge3\)

=> Min(*) = 3 <=> x + 1 = 0 => x = -1

=> MinM = \(2+\frac{52}{\left(-1+1\right)^2+3}=2+\frac{52}{3}=\frac{58}{3}\), đạt được khi x = -1

Mình không chắc nha -.-

\(M=\frac{2x^2+4x+60}{x^2+2x+4}=\frac{2\left(x^2+2x+4\right)+52}{x^2+2x+4}=2+\frac{52}{x^2+2x+4}\)

Để M đạt GTLN  => \(\frac{52}{x^2+2x+4}\)(**) đạt GTLN 

Hay \(x^2+2x+4\)(*) đạt GTNN 

Ta có : \(x^2+2x+4=\left(x^2+2x+1\right)+3=\left(x+1\right)^2+3\)

Do \(\left(x+1\right)^2\ge0\forall x\Leftrightarrow\left(x+1\right)^2+3\ge3\forall x\)

Nên GTNN (*) = 3 khi x + 1 = 0 <=> x = -1

Suy ra GTLN (**) = 52/3 khi x = -1

Vậy nên GTLN M = 2 + 52/3 = 58/3 khi x = -1

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

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 

\(y=5\left[\dfrac{3}{5}sin\left(3x+\dfrac{\pi}{6}\right)+\dfrac{4}{5}cos\left(3x+\dfrac{\pi}{6}\right)\right]\)

\(y=5.sin\left(3x+\dfrac{\pi}{6}+a\right)\) với \(cosa=\dfrac{3}{5}\)

Do \(-1\le sin\left(3x+\dfrac{\pi}{6}+a\right)\le1\)

\(\Rightarrow-5\le y\le5\)

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

Có điều kiện không??

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