Ta có:

\(B=-5x^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9x^2\)

\(=\left(2x-1\right)^2-\left(3x\right)^2\)

\(=\left(2x-1+3x\right)\left(2x-1-3x\right)\)

\(=-\left(x+1\right)\left(5x-1\right)\)

\(B=-5x^2-4x+1\)

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

\(B=-5\left[x^2+2.x.\frac{2}{5}+\left(\frac{2}{5}\right)^2-\frac{9}{25}\right]\)

\(B=-5\left(x+\frac{2}{5}\right)^2+5.\frac{9}{25}\)

\(B=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\)

Ta có: \(\left(x+\frac{2}{5}\right)^2\ge0\forall x\)

\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2\le0\forall x\)

\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\forall x\)

\(B=\frac{9}{5}\Leftrightarrow-5.\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)

Tham khảo nhé~

\(B=-5x^2-4x+1\)

\(=-5x^2-4x-\frac{4}{5}+\frac{9}{5}\)

\(=\left(-5x^2-4x-\frac{4}{5}\right)+\frac{9}{5}\)

\(=-5\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{9}{5}\)

\(=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)(vì \(-5\left(x+\frac{2}{5}\right)^2\le0\forall x\))

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

Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)

\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

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

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

Bài 2: 

a: Ta có: \(x^2+4x+7\)

\(=x^2+4x+4+3\)

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

giá trị nhỏ nhất chứ hình như sai đề

\(giải:\)

\(-4x^2+5x+1\)

\(=-4x^2+5x-\frac{25}{16}+\frac{41}{16}\)

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

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

\(=-\left[\left(2x\right)^2-2.2x.\frac{5}{4}+\left(\frac{5}{4}\right)^2\right]+\frac{41}{16}\)

\(=-\left(2x-\frac{5}{4}\right)^2+\frac{41}{16}\le\frac{41}{16}\)

\(GTLN\) \(của\)\(-4x^2+5x+1=\frac{41}{16}\)\(đạt\)\(khi\)\(-\left(2x-\frac{5}{4}\right)^2=0\)

                                                                                          \(\Leftrightarrow2x-\frac{5}{4}=0\)

                                                                                          \(\Leftrightarrow2x=\frac{5}{4}\Leftrightarrow x=\frac{5}{8}\)

vậy gtln của -4x^2+5x+1 bằng 41/16 tại x=5/8

Bài này giá trị lớn nhất nhé.

\(A=-4x^2+5x+1\)

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

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

\(A=-\left(\left(2x-\frac{5}{4}\right)^2-\frac{41}{16}\right)\)

\(A=-\left(2x-\frac{5}{4}\right)^2+\frac{41}{16}\le\frac{41}{16}\)

Vậy \(MaxA=\frac{41}{16}\Leftrightarrow-\left(2x-\frac{5}{4}\right)^2=0\)

                                  \(\Leftrightarrow x=\frac{5}{8}\)

A = - 4\(x\)² + 5\(x\) - 3

A = -( 4\(x^2\) - 5\(x\) + \(\dfrac{25}{16}\))  - \(\dfrac{23}{16}\)

A = -( 2\(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\)

Vì ( 2\(x\) - \(\dfrac{5}{4}\))² ≥ 0; ⇒ - ( 2\(x\) - \(\dfrac{5}{4}\))² ≤ 0 ⇒ -( 2 \(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\) ≤ - \(\dfrac{23}{16}\)

A(max) = - \(\dfrac{23}{16}\) ⇔ 2\(x\) - \(\dfrac{5}{4}\) = 0 ⇔ \(x\) = \(\dfrac{5}{4}\): 2 = \(\dfrac{5}{8}\)

Kết luận giá trị lớn nhất của biểu thức là - \(\dfrac{23}{16}\) xáy ra khi \(x\) = \(\dfrac{5}{8}\)

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

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

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