\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

         \(A=\frac{x^2}{x^4+x^2+1}\)

\(\Rightarrow\)\(3A=\frac{3x^2}{x^4+x^2+1}=\frac{x^4+x^2+1-x^4+2x^2-1}{x^4+x^2+1}\)

                 \(=\frac{\left(x^4+x^2+1\right)-\left(x^2-1\right)^2}{x^4+x^2+1}=1-\frac{\left(x^2-1\right)^2}{x^4+x^2+1}\le1\) 

\(\Rightarrow\)\(A\le\frac{1}{3}\)

Dấu  "=" xảy ra  \(\Leftrightarrow\)\(x=\pm1\)

Vậy  Max A = 1/3  <=>  \(x=\pm1\)

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

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

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

a) ĐKXĐ \(x +1\ne0=>x\ne-1;2x-6\ne0=>x\ne3\)

b) ta có 

\(P=\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\frac{3x}{2x-6}\)

để P = 1 thì \(\frac{3x}{2x-6}=1= >3x=2x-6\)

                                                        \(< =>3x-2x=-6=>x=-6\)

Max : với x = 0 thì \(A=\frac{x^2}{x^4+x^2+1}=0\)

với x khác 0 thì x⁴ + 1 \(\ge\)2x² > 0 nên x⁴ + x² + 1 \(\ge\)3x² 

\(\Rightarrow\)\(A=\frac{x^2}{x^4+x^2+1}\le\frac{x^2}{3x^2}=\frac{1}{3}\)

Vậy max A = \(\frac{1}{3}\)\(\Leftrightarrow\)x = 1 hoặc -1

Min : Ta có : x⁴ + x² + 1 = ( x²+ 1 )² - x² = ( x² - x + 1 ) ( x² + x + 1 ) > 0 

\(\Rightarrow\)\(A\ge0\)( vì x² \(\ge\)0 )

^x2-3.(x-1)

(x-1)²

^=>x2-3

x-1

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

Vậy GTNN của A là \(-\dfrac{4}{9}\) khi x = \(-\dfrac{2}{9}\)

\(B=25x^2+x-1=\left(25x^2+x+\dfrac{1}{100}\right)-\dfrac{101}{100}=\left(5x+\dfrac{1}{10}\right)^2-\dfrac{101}{100}\ge-\dfrac{101}{100}\)

Vậy GTNN của B là \(-\dfrac{101}{100}\) khi x = \(-\dfrac{1}{50}\)

\(C=3x^2+4x+1=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{1}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{1}{3}\ge-\dfrac{1}{3}\)

Vậy GTNN của C là \(-\dfrac{1}{3}\) khi x = \(-\dfrac{2}{3}\)

thanks

P= -(x² + 2.3x/2 + 9/4) +3 +9/4

GTLN: P = 21/4

\(P=-\left(x^2+3x-3\right)=-\left(x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{21}{4}\right)=-\left[\left(x+\frac{3}{2}\right)^2-\frac{21}{4}\right]=-\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\)

Do \(\left(x+\frac{3}{2}\right)^2\ge0,x\in R\)

nên \(-\left(x+\frac{3}{2}\right)^2\le0,x\in R\) 

mà \(-\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\le\frac{21}{4},x\in R\)

VẬy \(Max_P=\frac{21}{4}\)khi \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)

a/ Để biểu thức có nghĩa thì: \(\hept{\begin{cases}2x-2\ne0\\2-2x^2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)

b/ \(C=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)

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

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

c/ Có: \(C=-\frac{1}{2}\Leftrightarrow\frac{1}{2\left(x+1\right)}=-\frac{1}{2}\Rightarrow\frac{1}{x+1}=-1\)

               \(\Rightarrow-x-1=1\Rightarrow-x=2\Rightarrow x=-2\)

          Vậy x = -2

\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

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

Áp dụng BĐT Bunhiacopxki:

\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)

\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

\(\Rightarrow\left(x+y+z\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)