\(A=-3\left(x^2-\dfrac{5}{3}x+\dfrac{1}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{13}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{13}{12}< =\dfrac{13}{12}\)

Dấu = xảy ra khi x=5/6

Câu 1:

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

\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)

\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

            Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

    Vậy Min M = 11/4 khi x=3/2

b)\(N=2x^2+3x\)

\(N=2\left(x^2+\frac{3}{2}x\right)\)

\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)

\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

              Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)

                       Vậy MIn N = -9/8 khi x=-3/4

c)Tự làm nha

Ta có : x² - 3x + 5 

= x² - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)

= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)

Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)

Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)

Câu 2:

a)\(A=-x^2-5x+3\)

\(A=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{37}{4}\)

\(A=\frac{37}{4}-\left(x+\frac{5}{2}\right)^2\le\frac{37}{4}\)

            Dấu = xảy ra khi \(x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)

                      Vậy Max A = 37/4 khi x=-5/2

b)\(B=-2x^2+3x\)

\(B=-2\left(x^2-\frac{3}{2}x\right)\)

\(B=-2\left(x^2-2.\frac{3}{4}+\frac{9}{16}\right)+\frac{9}{8}\)

\(B=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)

         Dấu = xảy ra khi \(x-\frac{3}{4}=0\Rightarrow x=\frac{3}{4}\)

                    Vậy Max B=9/8 khi x=3/4

Sửa chút đề nhé! 

Với x khác -5/3

A= (3x^3+5x^2-9x-15):(3x+5)

= [x^2(3x+5)-3(3x+5)]:(3x+5)

 =(x^2-3) (3x+5):(3x+5)

=x^2-3\(\ge-3\)

Dấu '=' xảy ra khi x=0

max A=-3 khi x=0

\(-3x^2-9x+6\)

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

\(=-3\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{17}{4}\right)\)

\(=-3\left(x+\dfrac{3}{2}\right)^2+\dfrac{51}{4}\le\dfrac{51}{4}\forall x\)

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

\(A=\frac{2}{-5x^2+3x+2}=\frac{2}{\left(-5x^2+3x-\frac{9}{20}\right)+\frac{49}{20}}\)

\(A=\frac{2}{-5\left(x^2-\frac{3}{5}+\frac{9}{100}\right)+\frac{49}{20}}=\frac{2}{-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}}\ge\frac{2}{\frac{49}{20}}=\frac{40}{49}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-5\left(x-\frac{3}{10}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{3}{10}\)

Vậy GTNN của \(A\) là \(\frac{40}{49}\) khi \(x=\frac{3}{10}\)

\(B=\frac{5}{5x^2+4x+1}=\frac{5}{\left(5x^2+4x+\frac{4}{5}\right)+\frac{1}{5}}\)

\(B=\frac{5}{5\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{1}{5}}=\frac{5}{5\left(x+\frac{2}{5}\right)^2+\frac{1}{5}}\le\frac{5}{\frac{1}{5}}=25\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(5\left(x+\frac{2}{5}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{-2}{5}\)

Vậy GTLN của \(B\) là \(25\) khi \(x=\frac{-2}{5}\)

Chúc bạn học tốt ~ 

a) Ta có: A bé nhất khi \(-5x^2+3x+2\) lớn nhất

Ta có: \(-5x^2+3x+2=\left(-5x^2+3x-\frac{9}{20}\right)+\frac{49}{20}\)

\(=-5\left(x^2-2.\frac{3}{10}+\frac{9}{100}\right)=-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}\le\frac{49}{20}\)

Do đó \(A=\frac{2}{-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}}\le\frac{40}{49}\)

Dấu "=" xảy ra \(\Leftrightarrow-5\left(x-\frac{3}{10}\right)^2=0\Leftrightarrow x=\frac{3}{10}\)

Vậy \(A_{max}=\frac{40}{49}\Leftrightarrow x=\frac{3}{10}\)

b) Để B lớn nhất thì \(5x^2+4x+1\) bé nhất.Ta có:

\(5x^2+4x+1=\left(5x^2+4x\right)+1\)

\(=5\left(x^2+\frac{4}{5}x\right)+1=5\left(x^2+2.\frac{4}{10}+\frac{4}{25}\right)+\frac{1}{5}\)

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

Do đó \(B=\frac{5}{5\left(x+\frac{2}{5}\right)^2}\le\frac{5}{\frac{1}{5}}=25\)

Dấu "=" xảy ra \(\Leftrightarrow5\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy \(B_{max}=25\Leftrightarrow x=-\frac{2}{5}\)

A = 3x² - 7x + 8 = 3(x² - 7/3 + 49/36) + 47/12 = 3(x - 7/6)² + 47/12

Ta luôn có: 3(x - 7/6)² \(\ge\)0 \(\forall\)x

=> 3(x - 7/6)² + 47/12 \(\ge\)47/12 \(\forall\)x

Dấu "=" xảy ra khi: x - 7/6 = 0 <=> x = 7/6

Vậy Min của A = 47/12 tại x = 7/6

A= (x+5/2)^2 + 3/4 
nx 
(x+5/2)^2 >=0 suy ra a>= 3/4 
dấu bằng xảy ra khi x+5/2 = 0 <-> x=-5/2 

lop 8 con lic le 3