\(xy+2y=x+4\Leftrightarrow x\left(y-1\right)=4-2y\Leftrightarrow x=\frac{4-2y}{y-1}\left(y\ne1\right)\) 

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

Để x nguyên \(\Rightarrow\frac{2}{y-1}\) nguyên hay y-1 là ước của 2

\(\Rightarrow y-1=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow y=\left\{-1;0;2;3\right\}\Rightarrow x=\left\{-3;-4;0;-1\right\}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(\left(x^2+4x+x+4\right)\left(x^2+2x+3x+6\right)-24\)

\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+4=a\) ta có

\(a.\left(a+2\right)-24\)

\(a^2+2a-24\)

\(a^2+6a-4a-24\)

\(a\left(a+6\right)-4\left(a+6\right)\)

\(\left(a+6\right)\left(a-4\right)\)

\(\left(x^2+5x+4+6\right)\left(x^2+5x+4-4\right)\)

\(\left(x^2+5x+10\right)\left(x^2+5x\right)\)

   \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

   Đặt \(x^2+5x+5=a\)

   Suy ra  \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

               \(=\left(a+1\right)\left(a-1\right)-24\)

                 \(=a^2-1-24=a^2-25=\left(a-5\right)\left(a+5\right)\)

  Do đó 

         \(\left(a+5\right)\left(a-5\right)=x\left(x^2+5x+10\right)\left(x+5\right)\)

               Vậy \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=x\left(x^2+5x+9\right)\left(x+5\right)\)

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

\(P=\left[\left(x^2+1\right)^2-x^2\right]\left(x^4-x^2+1\right)-x^8-x^4\)

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

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

\(P=1\)

c\(=\)60⁰

cho lời giải giùm cái đi

a³ + b³ + c³ = 3abc

<=> a³ + b³ + c³ - 3abc = 0

<=> ( a³ + b³ ) + c³ - 3abc = 0

<=> ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

<=> [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

<=> ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

<=> ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

<=> ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

+) a + b + c = 0 => \(\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

=> \(M=\left(a+b\right)\left(b+c\right)\left(a+b\right)+abc=-abc+abc=0\)

+) a² + b² + c² - ab - bc - ac = 0

<=> 2( a² + b² + c² - ab - bc - ac ) = 2.0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

<=> ( a - b )² + ( b - c )² + ( a - c )² = 0

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Leftrightarrow a=b=c\)( chỗ này bạn tự đánh giá nhé )

=> \(M=\left(a+b\right)\left(b+c\right)\left(a+c\right)+abc=\left(a+a\right)\left(a+a\right)\left(a+a\right)+a\cdot a\cdot a\)

\(=\left(2a\right)^3+a^3=8a^3+a^3=9a^3\)( chỗ này thay là a, b, c tùy bạn nhé ; bằng nhau mà :)) )

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