\(x^2-4x+y+4\)

\(=\left(x^2-4x+4\right)+y\)

\(=\left(x-2\right)^2+y\)

\(x^3+3x^2-2xy+y=40\)

\(\Leftrightarrow x^3+3x^2-40=y\left(2x-1\right)\)

\(\Leftrightarrow y=\frac{x^3+3x^2-40}{2x-1}\)

\(y\inℤ\)suy ra \(\frac{x^3+3x^2-40}{2x-1}\inℤ\Rightarrow\frac{8\left(x^3+3x^2-40\right)}{2x-1}=\frac{8x^3+24x^2-320}{2x-1}=4x^2+14x-7-\frac{313}{2x-1}\inℤ\)

\(\Leftrightarrow\frac{313}{2x-1}\inℤ\Leftrightarrow2x-1\inƯ\left(313\right)=\left\{1,313\right\}\)(vì \(x\)nguyên dương) 

\(\Leftrightarrow x\in\left\{1,157\right\}\)

\(x=1\Rightarrow y=-36\left(l\right),x=157\Rightarrow y=12600\left(tm\right)\)

\(x^2-5x+6\)

\(=\left(x^2-2x\right)-\left(3x-6\right)\)

\(=x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(x-2\right)\left(x-3\right)\)

HT

\(\left(5x^3+14x^2+12x+8\right):\left(x+2\right).\)

\(=\left(5x^3+10x^2+4x^2+8x+4x+8\right):\left(x+2\right)\)

\(=\left[5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left(x+2\right).\left(5x^2+4x+4\right):\left(x+2\right)\)

\(=5x^2+4x+4\)

bằng 1

( 8x – 1 ) ( 3x + 2) + ( 6x + 1 )( 3 – 4x) = 28

<=> 24x + 16x - 3x -2 + 18x - 24x + 3 - 4x =28

<=> 24x -24x +16x - 3x + 18x - 4x = 2 - 3 + 28 

<=> 27x = 27

<=> x=27/27

<=> x=1 

Vậy x=1

mình hơi ẩu làm lại :(

( 8x – 1 ) ( 3x + 2) + ( 6x + 1 )( 3 – 4x) = 28

=> 24x^2 + 16x - 3x - 2 + 18x - 24x^2 + 3 - 4x =28

=> 24x^2 - 24x^2 + 16x -3x+18x-4x = 2-3+28

=> 27x = 27

=> x=27/27

=> x=1

Vậy x=1