=-1/2x^2+5x^2y^3-8x^3y^2-5x^2y^3+7x^3y^2-6x^2-5/3y

=(-1/2x^2+6x^2)+(5x^2y^3-5x^2y^3)+(-8x^3y^2-7x^3y^2)+5/3y

=11/2x^2+0-15x^3y^2+5/3y

=11/2x^2-15x^3y^2+5/3y

thay x=-1/2 , y=25 vào giá trị biểu thức M ta đc

       11/2.(-1/2)^2-15.(-1/2)^3.25^2+5/3.25=7273/6

   vậy tại x=-1/2 , y=25 vào giá trị biểu thức M có giá trị là 7273/6

Thay x = \(\dfrac{1}{2}\) vào biểu thức M ta có :

\(\dfrac{5.(\dfrac{1}{2})^2-7.\dfrac{1}{2}+1}{3.\dfrac{1}{2}-1}\) = \(\dfrac{\dfrac{-5}{4}}{\dfrac{1}{2}}\) = \(\dfrac{-5}{8}\)

Thay x=\(\dfrac{-1}{2}\) vào biểu thức M ta có :

\(\dfrac{5.(\dfrac{-1}{2})^2-7.(\dfrac{-1}{2})+1}{3.(\dfrac{-1}{2})-1}\) \(= \dfrac{\dfrac{23}{4}}{\dfrac{-5}{2}}\) \(= \dfrac{-115}{8}\)

\(ĐKXĐ:\)\(x\ne\left\{0;1;2;3;4;5\right\}\)

\(P=\frac{1}{x^2-x}+\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}\)

\(=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}\)

\(=\frac{1}{x-5}-\frac{1}{x}\)

\(=\frac{5}{x\left(x-5\right)}\)

Ta có:     \(x^3-x^2+2=0\)

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

Xét:    \(x^2-2x+2=\left(x-1\right)^2+1\)\(>0\)

\(\Rightarrow\)\(x+1=0\)

\(\Leftrightarrow\)\(x=-1\)(t/m)

Vậy   tại     \(x=-1\)  thì:

          \(P=\frac{5}{-1\left(-1-5\right)}=\frac{5}{6}\)

ĐKXĐ \(x\ne0,1,2,3,4,5\)

\(P=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

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

\(P=\frac{1}{x-5}-\frac{1}{x}\)

\(P=\frac{5}{x\left(x-5\right)}\)

a) \(P=\dfrac{2x-4}{x^2-4x+4}-\dfrac{1}{x-2}=\dfrac{2\left(x-2\right)}{\left(x-2\right)^2}-\dfrac{1}{x-2}\)

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

ĐKXĐ: \(x\ne2\) nên với x = 2 thì P không được xác định

\(Q=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3\left(x+5\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3x+15+x-3-2\left(x+3\right)}{x^2-9}=\dfrac{2x+6}{x^2-9}=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x-3}\)

Tại x = 2 thì \(Q=\dfrac{2}{2-3}=\dfrac{2}{-1}=-2\)

b) Để P < 0 tức \(\dfrac{1}{x-2}< 0\) mà tứ là 1 > 0

nên để P < 0 thì x - 2 < 0 \(\Leftrightarrow x< 2\)

Vậy x < 2 thì P < 0

c) Để Q nguyên tức \(\dfrac{2}{x-3}\) phải nguyên

mà \(\dfrac{2}{x-3}\) nguyên khi x - 3 \(\inƯ_{\left(2\right)}\)

hay x - 3 \(\in\left\{-2;-1;1;2\right\}\)

Lập bảng :

x - 3 -1 -2 1 2

x 2 1 4 5

Vậy x = \(\left\{1;2;4;5\right\}\) thì Q đạt giá trị nguyên

a) \(\dfrac{20x^3}{11y^2}.\dfrac{55y^5}{15x}=\dfrac{20.5.11.x.x^2.y^2.y^3}{11.3.5.x.y^2}=\dfrac{20x^2y^3}{3}\)

b) \(\dfrac{5x-2}{2xy}-\dfrac{7x-4}{2xy}=\dfrac{5x-2-7x+4}{2xy}=\dfrac{-2x+2}{2xy}=\dfrac{2\left(1-x\right)}{2xy}=\dfrac{1-x}{xy}\)