x² + y² + z² = xy + yz + zx

=> 2(x² + y² + z²) = 2 (xy + yz + zx)

=> 2x² + 2y² + 2z² = 2xy  + 2yz + 2zx

=> 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

=> (x² - 2xy + y²) + (y² - 2yz + z²) + (x² - 2xz + z²) = 0

=> (x - y)² + (y - z)² + (z - x)² = 0

=> \(\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}}\Rightarrow x=y=z\)

Khi đó x²⁰²⁰ + y²⁰²⁰ + z²⁰²⁰ = 3

<=> x²⁰²⁰ + x²⁰²⁰ + x²⁰²⁰ = 3 (Vì x = y = z)

=> 3.x²⁰²⁰ = 3

=> x²⁰²⁰ = 1

=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Khi x = 1 => A = x²⁰¹⁹ = 1²⁰¹⁹ = 1

Khi x = -1 => A = x²⁰¹⁹ = (-1)²⁰¹⁹ = -1

( x + 1 )⁴ + ( x - 3 )⁴ = 82

Đặt t = x - 1

pt <=> ( t + 2 )⁴ + ( t - 2 )⁴ = 82

<=> t⁴ + 8t³ + 24t² + 32t + 16 + t⁴ - 8t³ + 24t² - 32t + 16 - 82 = 0

<=> 2t⁴ + 48t² - 50 = 0

Đặt a = t²

<=> 2a² + 48a - 50 = 0

<=> 2a² - 2a + 50a - 50 = 0

<=> 2a( a - 1 ) + 50( a - 1 ) = 0

<=> ( a - 1 )( 2a + 50 ) = 0

<=> ( t² - 1 )( 2t² + 50 ) = 0

<=> ( t - 1 )( t + 1 )( 2t² + 50 ) = 0

<=> ( x - 1 - 1 )( x - 1 + 1 )[ 2( x - 1 )² + 50 ] = 0

<=> x( x - 2 )[ 2( x - 1 )² + 50 ] = 0

Vì 2( x - 1 )² + 50 ≥ 50 > 0 ∀ x

=> x( x - 2 ) = 0

<=> x = 0 hoặc x = 2

Vậy phương trình có hai nghiệm x = 0 hoặc x = 2 

(x + 1)⁴ + (x - 3)⁴ = 82

=> (x + 1)⁴ + (3 - x)⁴ = 82

=> x⁴ + 4x³ + 6x² + 4x + 1 + 81 + 108x - 54x² - 12x³ + x⁴ = 82

=> 2x⁴ - 8x³ - 48x² + 112x = 0

=> x⁴ - 4x³ - 24x² + 56x = 0

=> x(x³ - 4x² - 24x + 56) = 0

=> x(x³ - 4x² + 4x - 28x + 56) = 0

=> x[x(x² - 4x + 4) - 28(x - 2)] = 0

=> x[x(x - 2)² - 28(x - 2)] = 0

=> x(x - 2)[x(x - 2) - 28] = 0

=> x(x - 2)(x² - 2x - 28] = 0

=> x(x - 2)(x² - 2x + 1 - 29) = 0

=> x(x - 2)[(x - 1)² - 29] = 0

=> \(x\left(x-2\right)\left(x-1+\sqrt{29}\right)\left(x-1-\sqrt{29}\right)=0\)= 0

=> x = 0 hoặc x - 2 = 0 hoặc x - 1 + \(\sqrt{29}\)= 0 hoặc x - 1 -  \(\sqrt{29}\) = 0

=> x = 0 hoặc x = 2 hoặc x = - \(\sqrt{29}\)+ 1 hoặc x =  \(\sqrt{29}\)+ 1 

\(\frac{x^2-5x+1}{2x+1}+2=-\frac{x^2-4x+1}{x+1}\)

\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+\frac{4x+2}{2x+1}=-\frac{x^2-4x+1}{x+1}\)

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

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

\(\Leftrightarrow x^3+x^2-x^2-x+3x+3=\left(2x^3+x^2-8x^2-4x-2x-1\right)\)

rút gọn lại ... tự làm nốt nhé 

AA',BB',CC' là gì vậy bạn ?

\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)

Sai đề ở chỗ \(\left(\frac{x^2-16}{x-4}+1\right)\)thành -1