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

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

\(=-\left[\left(\dfrac{1}{2}x-1\right)^2+2\right]=-\left(\dfrac{1}{2}x-1\right)^2-2\le-2\)

\(\Rightarrow\)Biểu thức trên luôn âm ( đpcm )

a, \(x^2y^2.\sqrt{\dfrac{9}{x^2y^4}}=x^2y^2.\dfrac{3}{xy^2}=3x\)

b, \(\sqrt{\dfrac{x^2-6x+9}{x-3}}=\sqrt{\dfrac{\left(x-3\right)^2}{x-3}}=\sqrt{x-3}\)

\(A=x^2+14x+y^2-2y+7\)

\(=\left(x^2+14x+49\right)+\left(y^2-2y+1\right)-43\)

\(=\left(x+7\right)^2+\left(y-1\right)^2-43\ge-43\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(x+7\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=1\end{matrix}\right.\)

Vậy \(MIN_A=-43\) khi x = -7 và y = 1

Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)

\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)

\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)

\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)

Mà \(3a^2-5a-1=0\)

\(\Rightarrow f\left(a\right)=2011\)

Vậy...

\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{2013.2015.2017}\)

\(=\dfrac{1}{4}\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{2013.2015.2017}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}-\dfrac{1}{2015.2017}\right)\)\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{2015.2017}\right)=\dfrac{1}{12}-\dfrac{1}{4.2015.2017}\)

1, \(A=-x^2-2x-4=-\left(x^2+2x+1+3\right)\)

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

\(\Rightarrow\)A luôn âm

2, tương tự

3, \(C=-x^2-x-1=-\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

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

\(\Rightarrow\)C luôn âm

4, \(D=-2x^2+6x-8=-2\left(x^2-3x+4\right)\)

\(=-2\left(x^2-\dfrac{3}{2}.x.2+\dfrac{9}{4}+\dfrac{7}{4}\right)\)

\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{2}\le\dfrac{-7}{2}\)

\(\Rightarrow D\) luôn âm

5, tương tự

\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{2013.2015.2017}\)

\(=\dfrac{1}{4}\left(\dfrac{4}{1.2.3}+\dfrac{4}{3.5.7}+...+\dfrac{4}{2013.2015.2017}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}-\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{8}-\dfrac{1}{4.2015.2017}\)

1, Ta có: \(\left\{{}\begin{matrix}\left(3x-5\right)^{100}\ge0\\\left(2y-1\right)^{200}\ge0\end{matrix}\right.\Rightarrow\left(3x-5\right)^{100}+\left(2y-1\right)^{200}\ge0\)

Mà \(\left(3x-5\right)^{100}+\left(2y-1\right)^{200}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(3x-5\right)^{100}=0\\\left(2y-1\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy...

\(\dfrac{x+27}{343}+\dfrac{x+28}{342}=\dfrac{x+121}{249}+\dfrac{x+120}{250}\)

\(\Rightarrow\left(\dfrac{x+27}{343}+1\right)+\left(\dfrac{x+28}{342}+1\right)=\left(\dfrac{x+121}{249}+1\right)+\left(\dfrac{x+120}{250}+1\right)\)

\(\Rightarrow\dfrac{x+370}{343}+\dfrac{x+370}{342}-\dfrac{x+370}{249}-\dfrac{x+370}{250}=0\)

\(\Rightarrow\left(x+370\right)\left(\dfrac{1}{343}+\dfrac{1}{342}-\dfrac{1}{249}-\dfrac{1}{250}\right)=0\)

Mà \(\dfrac{1}{343}+\dfrac{1}{342}-\dfrac{1}{249}-\dfrac{1}{250}\ne0\)

\(\Rightarrow x+370=0\Rightarrow x=-370\)

Vậy x = -370

\(\left(x+6\right)\left(3x-1\right)+\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(3x-1+1\right)=0\)

\(\Leftrightarrow\left(x+6\right)3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

Vậy x = -6 hoặc x = 0