a, P = xy - 5x +20

Thay x=14 , y =5,5 vào đa thức P

P= 14.5,5 -5.14 +20

P= 14.(5,5-5) +20

P= 14. 0,5 +20

P= 7+20

P=27

Q= x²+xy-5x-5y

Thay x=-5 , y=-8

Q= (-5)²+(-5).(-8) -5.(-5)-5.(-8)

Q=(-5).[ (-5) +(-8) +5 +8 ]

Q= (-5). 0

Q= 0

Chúc bạn học tốt

P=xy -5x -4y+20 =x(y-5)-4(y-5)=(x-4)(y-5)

P(14,5,5) =(14-4)(5,5-5)=10.0,5=5

Q=x^2 +xy-5x-5y =x(x+y) -5(x+y) =(x+y)(x-5)

Q(-5,-8) =(-5-8)(-5-5) =130

1: x^2-9+3x(x-3)

=(x-3)(x+3)+3x(x-3)

=(x-3)(x+3+3x)

=(4x+3)(x-3)

2: =x^4+5x^2+4x+5

=x^4+x^3+x^2-x^3-x^2-x+5x^2+5x+5

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

3: =x^4+2x^2+1-x^2

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

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

1) \(x^2-9+3x\left(x-3\right)\)

\(=\left(x^2-9\right)+3x\left(x-3\right)\)

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

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

\(=\left(x-3\right)\left(4x+3\right)\)

2) \(x^4+5x^2+4x+5\)

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

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

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

3) \(x^4+x^2+1\)

\(=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

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

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

-x²-y²+8x+4y-21
=(-x²+8x-16)+(-y²+4y-4)-1
=-(x-4)²-(y-2)²-1
rồi đấy tự làm đi

Ta co/-5x/=-5x khi -5x <0=>x< 0

        /-5x/=5x khi-5x>0=>x>0

*-5x = 2x+21

<=>-5x-2x=21

<=>-7x=21

<=>3(loai)

*5x=2x-21

<=>5x-2x=21

<=>3x=21<=>x=3

toi cung dang dinh hoi

Lời giải:
a.

$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$

$=9x=9.15=135$

b.

$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$

$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$

c.

$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$

$=-8x^3+9x^2y^2-xy^3$

$=(-2x)^3+(3xy)^2-xy^3$

$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow8x^2\left(x+\frac{1}{2}\right)+8x\left(x+\frac{1}{2}\right)+2\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(8x^2+8x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\2\left(4x^2+4x+1\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\end{cases}}\)

Vậy pt có 1 No là...

\(2\left(x+5\right)-x^2-5x=0.\)

\(\Leftrightarrow2x+10-x^2-5x=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

nhầm 

\(\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

\(M=\left(5x-3y+3xy+x^2y^2\right)-\left(\dfrac{1}{2}x+2xy-y+4x^2y^2\right)\)

\(=5x-3y+3xy+x^2y^2-\dfrac{1}{2}x-2xy+y-4x^2y^2\)

\(=\left(5x-\dfrac{1}{2}x\right)+\left(y-3y\right)+\left(3xy-2xy\right)+\left(x^2y^2-4x^2y^2\right)\) \(=4,5x-2y+xy-3x^2y^2\)

Thay \(x=1;y=-\dfrac{1}{2}\) vào ta có:

\(4,5x-2y+xy-3x^2y^2\)

\(=4,5.1-2.\left(-\dfrac{1}{2}\right)+1.\left(-\dfrac{1}{2}\right)-3.1^2.\left(-\dfrac{1}{2}\right)^2\)

\(=4,5+1-\dfrac{1}{2}-\dfrac{3}{4}\) \(=\dfrac{17}{4}\)