a) Vậy A = 3/4 <=> x = -1/2 A = x 2 + x + 1 A = x 2 + 2. x + + 1 2 1 4 3 4 A = (x + ) 2 + ≥ 1 2 3 4 3 4

\(A=x^2-x+1\)

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

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

Dấu"=" xảy ra khi \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Vậy \(Min_A=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)

\(P=\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\)

ĐKXĐ : \(x\ne1\)

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

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

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

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

b) Xét hiệu P - 1/3 ta có :

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

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

Ta có : ( x - 1 )² ≥ 0 ∀ x => -( x - 1 )² ≤ 0 ∀ x

x² + x + 1 = ( x² + x + 1/4 ) + 3/4 = ( x + 1/2 )² + 3/4 ≥ 3/4 > 0 ∀ x

=> 3( x² + x + 1 ) ≥ 9/4 > 0 ∀ x

Vậy -( x - 1 )² và 3( x² + x + 1 ) trái dấu nhau

=> \(\frac{-\left(x-1\right)^2}{3\left(x^2+x+1\right)}\le0\)hay P - 1/3 ≤ 0

Đẳng thức xảy ra <=> x = 1 ( ktm ) => Không xảy ra đẳng thức

Vậy P < 1/3 ( đpcm )

\(P=\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\)

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

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

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

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

\(=\frac{1}{x+1}-\frac{1}{x}\)

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

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

b) \(\frac{2x+2y}{y-x}-\frac{x^2+xy}{3x^2-3y^2}=\frac{-2x-2y}{x-y}-\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}\)

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

\(=\frac{-2x-2y}{x-y}-\frac{x}{3\left(x-y\right)}\)

\(=\frac{3\left(-2x-2y\right)}{3\left(x-y\right)}-\frac{x}{3\left(x-y\right)}\)

\(=\frac{-6x-6y}{3\left(x-y\right)}-\frac{x}{3\left(x-y\right)}\)

\(=\frac{-7x-6y}{3\left(x-y\right)}\)

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

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

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

b, \(\frac{2x+2y}{y-x}-\frac{x^2+xy}{3x^3-3y^2}=-\frac{2x+2y}{x-y}-\frac{x^2+xy}{3x\left(x^2-y^2\right)}\)

\(=-\frac{2x+2y}{x-y}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)

\(=-\frac{6x\left(x+y\right)^2}{3x\left(x-y\right)\left(x+y\right)}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)

\(=-\frac{6x\left(x^2+2xy+y^2\right)}{3x\left(x-y\right)\left(x+y\right)}-\frac{x^2+xy}{3x\left(x-y\right)\left(x+y\right)}\)

\(=\frac{-12x^3-12x^2y-6xy^2-x^2-xy}{3x\left(x-y\right)\left(x+y\right)}\)

check hộ ý b nhá :)) 

Ta có:\(a^{2000}+b^{2000}=a^{2001}+b^{2001}=a^{2002}+b^{2002}\)

\(\Rightarrow a^{2000}+b^{2000}+a^{2002}+b^{2002}=2\left(a^{2001}+b^{2001}\right)\)

\(\Rightarrow a^{2002}-a^{2001}-a^{2001}+a^{2000}+b^{2002}-b^{2001}-b^{2001}+b^{2000}=0\)

\(\Rightarrow a^{2001}\left(a-1\right)-a^{2000}\left(a-1\right)+b^{2001}\left(b-1\right)-b^{2000}\left(b-1\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a^{2001}-a^{2000}\right)+\left(b-1\right)\left(b^{2001}-b^{2000}\right)=0\)

\(\Rightarrow a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)

Dấu"="xảy ra khi \(\orbr{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\)Mà \(a,b>0\)

\(\Rightarrow a=b=1\)

Do đó:\(a^{2020}+b^{2020}=1^{2020}+1^{2020}=1+1=2\)

Bài 1.

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

b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)

\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)

c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)

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

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

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

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)

Bài 3.

N = ( 4x + 3 )² - 2x( x + 6 ) - 5( x - 2 )( x + 2 )

= 16x² + 24x + 9 - 2x² - 12x - 5( x² - 4 )

= 14x² + 12x + 9 - 5x² + 20

= 9x² + 12x + 29

= 9( x² + 4/3x + 4/9 ) + 25

= 9( x + 2/3 )² + 25 ≥ 25 > 0 ∀ x 

=> đpcm