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\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)
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 \(x+2=0\Leftrightarrow x=-2\).
![](https://rs.olm.vn/images/avt/0.png?1311)
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~
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
1/
a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)
Dấu "=" xảy ra khi x=1/2
Vậy Amin=4 khi x=1/2
b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x=-1
Vậy Bmin = -4 khi x=-1
2/
a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)
Dấu "=" xảy ra khi x=3
Vậy Amax = 19 khi x=3
b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)
Dấu "=" xảy ra khi x=5/4
Vậy Bmax = 31/8 khi x=5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x
\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)
\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)
\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)
\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)
\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x
\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)
\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)
\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)
\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{4\left(x^2-4x+4\right)+\left(x^2-8x+16\right)}{x^2-4x+4}=4+\left(\dfrac{x-4}{x-2}\right)^2\ge4\)
\(A_{min}=4\) khi \(x=4\) (A max ko tồn tại)
\(B=\dfrac{6\left(x^2+2x+1\right)+\left(4x^2+12x+9\right)}{x^2+2x+1}=6+\left(\dfrac{2x+3}{x+1}\right)^2\ge6\)
\(B_{min}=6\) khi \(x=-\dfrac{3}{2}\)
B max ko tồn tại
![](https://rs.olm.vn/images/avt/0.png?1311)
a:Ta có: \(A=-4x^2+x-1\)
\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)
\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)
\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)
b: Ta có: \(B=-3x^2+5x+6\)
\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)
c: Ta có: \(C=-x^2+3x+4\)
\(=-\left(x^2-3x-4\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-4x+7=\left(x^2-4x+4\right)+3=\left(x-2\right)^2+3\)
Vì: \(\left(x-2\right)^2\ge0\)
=> \(\left(x-2\right)^2+3\ge3\)
Vậy GTNN của A là 3 khi x=2
\(B=2x^2+12x-1=2\left(x^2+6x+9\right)-19=2\left(x+3\right)^2-19\)
Vì: \(2\left(x+3\right)^2\ge0\)
=> \(2\left(x+3\right)^2-19\ge-19\)
Vậy GTNN của B là -19 khi x=-3
\(C=5x-x^2=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
Vì: \(-\left(x-\frac{5}{2}\right)^2\le0\)
=> \(-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Vậy GTLN của C là \(\frac{25}{4}\) khi \(x=\frac{5}{2}\)
Căm ơn bạn nhiều nhé ! Nếu được thì bạn làm giúp tớ bài hình bên trên nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2}{-5x^2+3x+2}=\frac{2}{\left(-5x^2+3x-\frac{9}{20}\right)+\frac{49}{20}}\)
\(A=\frac{2}{-5\left(x^2-\frac{3}{5}+\frac{9}{100}\right)+\frac{49}{20}}=\frac{2}{-5\left(x-\frac{3}{10}\right)^2+\frac{49}{20}}\ge\frac{2}{\frac{49}{20}}=\frac{40}{49}\)
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 \(B_{max}=25\Leftrightarrow x=-\frac{2}{5}\)