x^3 + (x - 2)(2x + 1) = 8

<=> x^3 + 2x^2 - 3x - 2 - 8 = 0

<=> x^3 + 2x^2 - 3x - 10 = 0

<=> (x - 2)(x^2 + 4x + 5) = 0

vì x^2 + 4x + 5 > 0 nên:

<=> x - 2 = 0

<=> x = 2

       x^3+(x-2)(2x+1)=8

<=>x^3+2x^2-3x-10=0

<=>x^3-2x^2+4x^2-8x+5x-10=0

<=>x^2(x-2)+4x(x-2)+5(x-2)=0

<=>(x-2)(x^2+4x+5)=0

Mà x^2+4x+5>0

=>x-2=0<=>x=2

   Hok tốt !

\(x+y=1\Rightarrow\hept{\begin{cases}x=1-y\\y=1-x\end{cases}}\)

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

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

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

\(A=\frac{\left(y-x\right)\left(x+y\right)+\left(y-x\right)}{x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(A=\frac{\left(y-x\right)\left(x+y+1\right)}{x^2y^2+x^2+y^2+xy\left(x+y\right)+xy+\left(x+y\right)+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\) mà x + y = 1

\(A=\frac{2\left(y-x\right)}{x^2y^2+x^2+y^2+2xy+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(A=\frac{2\left(y-x\right)}{x^2y^2+\left(x+y\right)^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\) ; x + y = 1

\(A=\frac{2\left(y-x\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

a)Thay x = 102 tmđk vào M ta có:

M=102^3-6.102^2+12.102-8

M=102^3-6.102^2+1224-8

M=102^3-6.10404+1216

M=102^3-62424+1216

M=1061208-62424+1216

M=1.000.000

Còn câu b mk ko bít làm :(((

M = x^2 + y^2 - 2x + 4y + 5

M = (x^2 - 2x + 1) + (y^2 + 4y + 4) 

M = (x - 1)^2 + (y + 2)^2

M = 0 <=> x - 1 = 0 và y + 2 = 0

<=> x = 1 và y = -2