giả thiết => \(\frac{M\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{N\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{32x-19}{\left(x+1\right)\left(x-2\right)}\)

=> M(x-2) + N(x+1) = 32x - 19

<=> M.x - 2.M + N.x + N = 32.x -19

=> (M+ N).x + (N - 2.M) = 32.x - 19

=> M+ N = 32 và -2M + N = -19 

=> M = 17, N = 15

vậy M.N = 17. 15 =...

a) \(x^3-4x^2+5x-2\)

=\(x^3-x^2-3x^2+3x+2x-2\)

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

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

=\(\left(x-1\right)\left(x^2-x-2x-2\right)\)

=\(\left(x-1\right)\left(\left(x\left(x-1\right)-2\left(x-1\right)\right)\right.\)

=\(\left(x-1\right)^2\left(x-2\right)\)

b) \(\) \(x^5+x+1\)

=\(x^5-x^2+x^2+x+1\)

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

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

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

c) \(x^3+5x^2+5x+1\)

=\(\left(x^3+1\right)+\left(5x^2+5x\right)\)

=\(\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)

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

d) \(x^2\left(x^2+2y^2\right)-3y^4\)

=\(x^4+2x^2y^2-3y^4\)

=\(x^4-x^2y^2+3x^2y^2-3y^4\)

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

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

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

ĐK: \(x\ne2\).

a) \(P=\frac{x+1}{x-2}=\frac{x-2+3}{x-2}=1+\frac{3}{x-2}\)nguyên mà \(x\)nguyên nên \(x-2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\)

suy ra \(x\in\left\{-1,1,3,5\right\}\).

Thử lại để \(P\)nguyên dương thì \(x\in\left\{-1,3,5\right\}\).

b) \(-x^2-x+2=0\)

\(\Leftrightarrow-x^2+x-2x+2=0\)

\(\Leftrightarrow\left(-x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\Rightarrow P=\frac{1}{4}\\x=1\Rightarrow P=-2\end{cases}}\)

tach phan nguyên nhí bn

\(C=\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

Dấu ''='' xảy ra khi \(x=1;y=-1\)

Vậy GTNN C là 0 khi x = 1 ; y = -1