a: -526,8<0

0<0,65

Do đó: -526,8<0,65

b: 6,45>0

0>-3,273

Do đó 6,45>-3,273

c: 7,78<9,56

=>-7,78>-9,56

d: 0,789>0,356

=>-0,789<-0,356

\(P\left(1\right)=1^3-2\cdot a\cdot1+a^2=a^2-2a+1\)

\(Q\left(3\right)=3^2+\left(3a+1\right)\cdot3+a^2=a^2+9a+12\)

P(1)=Q(3)

=>\(a^2+9a+12=a^2-2a+1\)

=>11a=-11

=>a=-1

x+y-3=0

=>x+y=3

\(M\left(x\right)=x^3+x^2y-3x^2-xy-y^2+4y+x+2020\)

\(=x^2\left(x+y\right)-3x^2-y\left(x+y\right)+4y+x+2020\)

\(=3x^2-3x^2-3y+4y+x+2020\)

=x+y+2020

=3+2020

=2023

C là trung điểm của AB

=>\(CA=CB=\dfrac{AB}{2}=5\left(cm\right)\)

O là trung điểm của AC

=>\(AO=\dfrac{AC}{2}=\dfrac{5}{2}=2,5\left(cm\right)\)

AC=5 cm ;  CB=5 cm ;  AO=2,5 cm

\(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)

\(=a^2b+ab^2-b^2c-bc^2+a^2c-ac^2\)

\(=\left(a^2b+a^2c\right)+\left(ab^2-ac^2\right)+\left(-b^2c-bc^2\right)\)

\(=a^2\left(b+c\right)+a\left(b-c\right)\left(b+c\right)-bc\left(b+c\right)\)

\(=\left(b+c\right)\left(a^2+ab-ac-bc\right)\)

\(=\left(b+c\right)\left[a\left(a+b\right)-c\left(a+b\right)\right]\)

=(a+b)(b+c)(a-c)

Theo bezout ta có:

\(x^2\) + a\(x\) + b : \(x\) + 1 dư 7

Khi và chỉ khi (-1)² + (-1).a + b = 7

                      1 - a + b = 7

                         b - a = 6

b: abc-(ab+bc+ca)+(a+b+c)-1

\(=abc-ab-bc-ca+a+b+c-1\)

\(=\left(abc-ab\right)+\left(c-1\right)+\left(-bc+b\right)+\left(-ca+a\right)\)

\(=ab\left(c-1\right)+\left(c-1\right)-b\left(c-1\right)-a\left(c-1\right)\)

\(=\left(c-1\right)\left(ab-b-a+1\right)\)

\(=\left(c-a\right)\left[b\left(a-1\right)-\left(a-1\right)\right]\)

=(a-1)(b-1)(c-a)

a: