ai giúp tớ vs

• Kết luận: \(x = -21\)

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 \(\frac{3x}{2x-6}=1= >3x=2x-6\)

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

4x(x²-2x+3)-3x(x+1)(x-2)+(2x-5)(x-7)-4x(x²-x+1)

=4x³-8x²+12x-3x(x²-2x+x-2)+2x²-14x-5x+35-4x³+4x²-4x

=4x³-8x²+12x-3x³+6x²-3x²+6x+2x²-14x-5x+35-4x³+4x²-4x

=-3x³+x²-5x+35

#H

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

         \(A=\frac{x^2}{x^4+x^2+1}\)

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

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

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

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/ 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

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 )