+) Ta có:

\(P=-3x^2-x+4=-\left(3x^2+x+\dfrac{3}{36}\right)+\dfrac{49}{12}\)

\(=-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{49}{12}\)

Vì: \(-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)^2\le0\forall x\Rightarrow-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)+\dfrac{49}{12}\le\dfrac{49}{12}\)

Dấu ''='' xảy ra khi \(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}=0\Leftrightarrow x=-\dfrac{1}{6}\)

Vậy \(MAX_P=\dfrac{49}{12}\Leftrightarrow x=-\dfrac{1}{6}\)

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

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

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

\(=-x^2+x+6\)

=>y'=-2x+1

Đặt y'=0

=>-2x+1=0

=>-2x=-1

=>\(x=\dfrac{1}{2}\)

\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+2\right)\left(3-\dfrac{1}{2}\right)=\dfrac{5}{2}\cdot\dfrac{5}{2}=\dfrac{25}{4}\)

\(f\left(-2\right)=\left(-2+2\right)\left(3+2\right)=0\)

\(f\left(3\right)=\left(3+2\right)\left(3-3\right)=0\)

=>\(y_{max\left[-2;3\right]}=\dfrac{25}{4}\)

1) Áp dụng BĐT Bunhiacopski

P = \(6\sqrt{x-1}+8\sqrt{3-x}\le\sqrt{\left(6^2+8^2\right)\left(x-1+3-x\right)}=10\sqrt{2}\)

Vậy Min P = \(10\sqrt{2}\) khi x = 43/25

2) a) \(\Rightarrow A-5=y-2x=4y.\dfrac{1}{4}+\left(-6x\right).\dfrac{1}{3}\)

Áp dụng BĐT bunhiacopski

\(\Rightarrow\left(A-5\right)^2=\left(4y.\dfrac{1}{4}+\left(-6x\right).\dfrac{1}{3}\right)^2\) \(\le\left(16y^2+36x^2\right)\left(\dfrac{1}{16}+\dfrac{1}{9}\right)=\dfrac{25}{16}\)

\(\Rightarrow-\dfrac{5}{4}\le A-5\le\dfrac{5}{4}\Rightarrow\dfrac{15}{4}\le A\le\dfrac{25}{4}\)

...........

b) tương tự

App giải toán không cần nhập đề chỉ cần chụp ảnh cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618