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

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

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

b) \(\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)

Để A đạt giá trị nguyên => \(\frac{2}{x+1}\)đạt giá trị nguyên

=> 2 ⋮ x + 1

=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }

So với ĐKXĐ ta thấy x = 0 ; x = -2 ; x = -3 thỏa mãn

Vậy x ∈ { -3 ; -2 ; 0 } thì A đạt giá trị nguyên

c) Tại x = -1/3 ( tm ) => A = \(\frac{-\frac{1}{3}-1}{-\frac{1}{3}+1}=-2\)

bài 1.a. điều kiện xác định của phân thức là \(x^3-8\ne0\Leftrightarrow x\ne2\)

b .ta có \(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x+2}\)

bài 2.

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

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

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

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

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

bài 1.

a.\(\left(x+4\right)\left(x^2-4x+16\right)=x^3-4^3=x^3-64\)

b.\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{27}\)

bài 2.

a.\(892^2+892.216+108^2=892^2+2.892.108+108^2\)

\(=\left(892+108\right)^2=1000^2=1_{ }000_{ }000\)

b.\(36^2+26^2-52.36=36^2+26^2-2.26.36=\left(36-26\right)^2=10^2=100\)

Tương tự mấy phần kia 

\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)

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

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

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

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

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

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

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

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

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

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

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

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

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