Đáp án:
1)
$2A-B=3x^2y+6xy^2+9x^2y^2+5x^4+1\\
A+B=12x^2y^2-2x^4-1\\
2)
P(x)=0
\Leftrightarrow x=2\\
P(x)=-|-5|
\Leftrightarrow x=\dfrac{11}{3}\\$
3)
A có ${\left[\begin{aligned}x=2\\x=3\end{aligned}\right.}$
B có $x=\dfrac{-2}{3}$
Giải thích các bước giải:
$1)
A=xy(x+2y) - x^2.(-7y^2 - x^2)\\
=x^2y+2xy^2+7x^2y^2+x^4\\
B= y^2.(5x^2 - 2x) - (yx^2 + 3x^4 + 1)\\
=5x^2y^2-2xy^2-x^2y-3x^4-1\\
2A-B=2.(x^2y+2xy^2+7x^2y^2+x^4)-(5x^2y^2-2xy^2-x^2y-3x^4-1)\\
=2x^2y+4xy^2+14x^2y^2+2x^4-5x^2y^2+2xy^2+x^2y+3x^4+1\\
=(2x^2y+x^2y)+(4xy^2+2xy^2)+(14x^2y^2-5x^2y^2)+(2x^4+3x^4)+1\\
=3x^2y+6xy^2+9x^2y^2+5x^4+1\\
A+B=(x^2y+2xy^2+7x^2y^2+x^4)+(5x^2y^2-2xy^2-x^2y-3x^4-1)\\
=x^2y+2xy^2+7x^2y^2+x^4+5x^2y^2-2xy^2-x^2y-3x^4-1\\
=(x^2y-x^2y)+(2xy^2-2xy^2)+(7x^2y^2+5x^2y^2)+(x^4-3x^4)-1\\
=12x^2y^2-2x^4-1\\
2)
P(x)=0\\
\Leftrightarrow 2x^3 + x^2 + 5 - 3x + 3x^2 -2x^3 - 4x^2 + 1=0\\
\Leftrightarrow (2x^3-2x^3)+(x^2+3x^2-4x^2)-3x+(5+1)=0\\
\Leftrightarrow -3x=-6\\
\Leftrightarrow x=2\\
P(x)=-|-5|=-5\\
\Leftrightarrow -3x+6=-5\\
\Leftrightarrow -3x=-11\\
\Leftrightarrow x=\dfrac{11}{3}\\
3)
A=(2-x)+(x-2)^2=0\\
\Leftrightarrow -(x-2)+(x-2)^2=0\\
\Leftrightarrow (x-2)(-1+x-2)=0
\Leftrightarrow (x-2)(x-3)=0\\
\Leftrightarrow {\left[\begin{aligned}x=2\\x=3\end{aligned}\right.}\\
B=\dfrac{1}{2}x^3+\dfrac{4}{27}=0\\
\Leftrightarrow \dfrac{1}{2}x^3=-\dfrac{4}{27}\\
\Leftrightarrow x^3=\dfrac{-8}{27}\\
\Leftrightarrow x^3=\left ( \dfrac{-2}{3} \right )^3\\
\Leftrightarrow x=\dfrac{-2}{3}$
